Tecl^nical Drawing Scries 



Descriptive Geometry 



ANTHONY AND ASHLEY 



I 




Copyright ]^^. 



COPYRIGHT DEPOSnV 



Technical Drawing Series 



ANTHONY'S MECHANICAL DRAWING 

ANTHONY'S MACHINE DRAWING 

ANTHONY'S GEARING 

ANTHONY AND ASHLEY'S DESCRIPTIVE GEOMETRY 

DANIELS'S FREEHAND LETTERING 

DANIELS'S TOPOGRAPHICAL DRAWING 



D. C. HEATH &L CO., PUBLISHERS 



TECHNICAL DRAWING SERIES 



DESCRIPTIVE GEOMETRY 



BY 
GARDNER C. ANTHONY, Sc.D. 

AND 

GEORGE F. ASHLEY 



BOSTON, U.S.A. 

D. C. HEATH & CO., PUBLISHERS 

1909 



Copyright, 1909, 
By D. C. Heath & Co. 



r\ 



LIBRARY of CONGRESS 

Tv/o Cociesi Received 

APR 3 1^09 
cUss «c '^»<=' ""■ 



PEEFACE 



Ax extended experience in engineering 
practice and teaching has emphasized the 
importance of certain methods for the presen- 
tation of principles and problems in graphics 
which the authors have gradually developed 
into their present form. Most of the subject- 
matter wliich we now publish has been pre- 
sented in the form of notes and used during 
the past eight years, and nearly all of the 
problems have received the critical test of the 
classroom. Previous to the preparation of 
the notes, whicli were made the foundation 
of this treatise, tlie third angle of projection 
had been adopted for problems, and almost 
exclusively used, as con form ijig more nearly 
to engineering practice. 

It has been the aim of the authors to make 
a clear and concise statement of the princi- 
ples involved, together Avith a brief analysis 
and enumeration of the steps to be taken, so 



that the essentials of each problem shall be 
clearly set forth. The illustrations have been 
chosen with care and arranged so as to appear 
opposite the descriptive text. 

Too much stress cannot be laid on tlie im- 
portance of preparing problems with such care 
that they may clearly illustrate the principles 
involved. In general it should not be neces- 
sary for the student to prescribe the condi- 
tions governing the relations of points, lines, 
and surfaces because of the time consumed 
and the probable failure to bring out the 
salient features of the problem. The graphic 
presentation of the problems to be solved 
should facilitate the lay-out by the student, 
and enable the instructor to judge quickly of 
their character and adaptability to the special 
cases under consideration. Two sets of prob- 
lems have been prepared to illustrate most 
cases, and the number may be further in- 



ui 



IV 



PREFACE 



creased by reversing, or inverting, the illus- 
trations given. The unit of measurement 
may also be changed to adapt the problems 
to any chosen size of plate, and if the propor- 
tion be maintained it will be possible to solve 
the problems within the given space. If it is 
desired to change the assignment annually, 
the even numbers may be used for one year 
and the odd numbers for the following year. 
It is hoped that the number of problems is 
sufficiently great to admit of considerable va- 
riety in the course. 



Tufts College, January, 1909. 



Although the authors believe that an ele- 
mentary course in the orthographic and iso- 
metric projections of simple objects should 
precede the more analytical consideration of 
line and plane, as herein presented, yet this 
treatise has been proven adequate to meet the 
demand of those who have not received such 
preparation. By thus making it available for 
students of science and mathematics, as well 
as for engineers, it is hoped to promote a more 
general knowledge of the grammatical con- 
struction of Graphic Language. 

^GARDNER C. ANTHONY 
GEORGE F. ASHLEY. 



TABLE OF CONTENTS 



1. Descriptive Geometry 

2. Projection 

3. Coordinate Planes . 

4. Quadrants or Angles 

5. Orthographic Projection 

6. Notation . 

7. Points 

8. Lines 

9. Line parallel to a coordin 
10. 



CHAPTER I 
Definitions and First Principles 



ate pi 



Line peipendicular to a coord 



ane 
nate plane 



1 IL Line lying in a coordinate plane 

1 12. Line parallel to a coordinate plane 

2 13. Lines parallel in space 

3 14. Lines intersecting in space 

3 15. Lines intersecting the ground line 

4 16. Traces of a line 

6 17. To define the position of a line 

7 18. Planes 

8 19. GL the trace of V and H 



PAGE 

8 
8 
8 
10 
10 
10 
10 
12 
12 



CHAPTER II 
Points, Lines, and Planes 



20. Operations required for the solution of prob- 

lems 

21. To determine three projections of a line 

22. Projection of point in 2 Q 

23. Revolving of P 

24. To determine the traces of a line . 

25. To determine the traces of a line parallel to P 

26. To determine the projections of a line when 

its traces are given . . . . . 



14 
14 
15 
15 

16 
18 

18 



Conditions governing lines lying in a plane . 19 
Conditions governing lines lying in a plane 

and parallel to H or V . . . . 19 

An infinite number of planes may be passed 

through any line 20 

To pass a plane through two intersecting or 

parallel lines, Case 1 20 

31. To pass a plane through two intersecting or 

parallel lines, Case 2 20 



27. 

28. 

29. 
30. 



VI 



TABLE OF CONTENTS 



AET. 

32. To pass a plane through two intersecting or 

parallel lines. Case 3 

33. To pass a plane through a line and a point . 

34. To pass a plane through three points not in 

the same straight line ..... 

35. Given one projection of a line lying on a plane, 

to determine the other projection 

36. Given one projection of a point lying on a 

plane, to determine the other projection 

37. To locate a point on a given plane at a given 

distance from the coordinate planes 

38. To revolve a point into either coordinate plane 

39. To determine the true length of a line, Case 1 

40. To determine the true length of a line. Case 2 

41. Relation of the revolved position of a line to 

its trace ., 

42. The revolved position of a point lying in a plane 

43. The revolved position of a line lying in a plane 

44. To determine the angle between two inter- 

secting lines 

45. To draw the projections of any polygon . 

46. Counter-revolution, Construction 1 . 

47. Counter-revolution, Construction 2 . 

48. Counter-revolution, Construction 3 . 

49. To determine the projections of the line of in- 

tersection between two planes. Principle . 

50. To determine the projections of the line of in- 

tersection between two planes. Case 1 . 

51. To determine the projections of the line of in- 

tersection between two planes. Case 2 

52. To determine the projections of the line of 

intei'section between two planes when two 
traces are parallel 



VGE 


AKT. 




53. 


21 




22 






54. 


22 






55. 


22 




24 


56. 


24 


57. 


25 




27 


58. 


28 






59. 


28 




30 


60. 


31 






61. 


31 




32 


62. 


32 




33 




34 


63. 




64. 


35 






65. 


35 






66. 


36 


67. 



37 



To determine the projections of the line of in- 
tersection between two planes when all 
traces meet in GL 37 

To determine the projections of the line of in- 
tersection between two planes, Case 3 . 38 

To determine the projections of the line of in- 
tersection between two planes when one 
plane contains GL 38 

Revolution, Quadrants, and Counter-revolu- 
tion 40 

To determine the point in which a line pierces 
a plane. General Method .... 42 

To determine the point in which a line pierces 
a plane. Case 1 42 

To determine the point in which a line pierces 
a plane. Case 2 42 

To determine the point in which a line pierces 
a plane. Case 3" 42 

To determine the point in which a line piei'ces 
a plane, Case 4^ 44 

If a right line is perpendicular to a plane, the 
projections of that line will be perpendicu- 
lar to the traces of the plane ... 44 

To project a point on to an obliqne plane . 44 

To project a given line on to a given oblique 
plane 45 

To determine the shortest distance from a 
point to a plane 45 

Shades and Shadows 46 

To determine the shadow of a point on a given 
surface 47 

To determine the shadow of a line npon a 
given surface . . . . . - 47 



TABLE OF CONTENTS 



Vll 



ART. PAGE 

69. To determine the shadow of a solid upon a 

given surface ...... 48 

70. Through a ]:>oint or line to pass a plane hav- 

ing a defined relation to a given line or 
plane 50 

71. To pass a plane througli a given point parallel 

to a given plane 50 

1'2. To pass a plane through a given point perpen- 
dicular to a given line 50 

73. To pass a plane through a given point parallel 

to two given lines ..... 51 

74. To pass a plane through a given line parallel 

to another given line 52 

75. To pass a plane through a given line perpen- 

dicular to a given plane . . . .5*2 

76. Special conditions and methods of Art. 70 . 52 

77. To pass a plane through a given point perpen- 

dicular to a given line. Special case . . 52 

78. To pass a plane through a given line perpen- 

dicular to a given plane. Special case . 53 

79. To determine the projections and true length 

of the line measuring the shortest distance 
between two I'ight lines not in the same 
plane 54 

80. To determine the angle between a line and a 

plane 54 



AKT. 

81. To determine the angle between a line and the 

coordinate planes 

82. To determine the projections of a line of defi- 

nite length passing through a given point 
and making given angles with the coordi- 
nate planes 

83. To determine the angle between two planes. 

Principle 

84. To determine the angle between two oblique 

planes 

85. To determine the angle between two oblique 

planes by perpendiculars .... 

86. To determine the angle between an inclined 

plane and either coordinate plane 

87. To determine the bevels for the correct cuts, 

the lengths of hip and jack rafters, and the 
bevels for the purlins for a hip roof . 

88. Given one trace of a plane, and the angle be- 

tween the plane and the coordinate plane, 
to determine the otlier trace. Case 1 . 

89. Given one trace of a plane, and the angle be- 

tween the plane and the coordinate plane, 
to determine the other trace. Case 2 . 

90. To determine tlie traces of a plane, knowing 

the angles which the plane makes with both 
coordinate planes 

91. Fourth construction for counter-revolution 



56 
57 
57 

58 
58 

60 

62 



62 
63 



CHAPTER III 
Generation and Classification of Surfaces 



92. jNIethod of generating surfaces 

93. Classification of Surfaces 



65 
65 



94. 



Kuled Surfaces 
Plane Surfaces 



65 
66 



Vlll 



TABLE OF CONTENTS 



96. Single-curved Surfaces 

97. Conical Surfaces . 

98. Cylindrical Surfaces 

99. Convolute Surfaces 



66 100. A Warped Surface 

66 101. Types of Warped Surfaces 

67 102. A Surface of Revolution 
67 103. Double-curved Surfaces . 



PAGE 

68 
68 
70 
70 



CHAPTER IV 



Tangent Planes 



101. Plane tangent to a single-curved surface 

105. One projection of a point on a single-curved 

surface being given, it is required to pass 
a plane tangent to the surface at the ele- 
ment containing the given point 

106. To pass a plane tangent to a cone and through 

a given point outside its surface 

107. To pass a plane tangent to a cone and parallel 

to a given line 

108. To pass a plane tangent to a cylinder and 

through a given point outside its surface . 

109. To pass a plane tangent to a cylinder and 

parallel to a given line .... 



72 



73 



74 



112. 



75 113 



75 



77 



76 



110. Plane tangent to a double-curved surface 

111. One projection of a point on the surface of a 

double-curved surface of revolution being 
given, it is required to pass a plane tangent 
to the surface at that point 

Through a point in space to pass a plane 
tangent to a given parallel of a double- 
curved surface of revolution 

To pass a plane tahgent to a sphere at a given 

point on its surface 

114. Through a givendine to pass planes tangent 

to a sphere 80 



78 



79 



80 



CHAPTER V 
Tntersection of Planes with Surfaces, and the Development of Surfaces 



115. To determine the intersection of any surface 

with any secant plane .... 

116. A tangent to a curve of intersection 

117. The true size of the cut section 

118. A right section 



82 


120. 


83 




83 


121. 


83 


122. 



The development of a surface ... 83 
To determine the intersection of a plane with 

a pyramid 83 

To develop the pyramid .... 84 
To determine the curve of intersection be- 
tween a plane and any cone ... 86 



TABLE OF CONTENTS 



IX 



123. To determine the development of any oblique 
cone ........ 

124:. To determine the curve of intersection be- 
tween a plane and any cylinder. 

12.5. To develop the cylinder .... 

126. The development of a cylinder when the axis 

is parallel to a coordinate plane 

127. To determine the curve of intersection be- 

tween a plane and a prism 



86 

88 
88 

91 

92 



128. To develop the prism .... 

129. The Helical Convolute .... 

130. To draw elements of the surface of the heli 

cal convolute 

131. To develop the helical convolute . 

132. To determine the curve of intersection be 

tween a plane and a surface of revolu 
tion 



93 
93 

94 
9.5 



96 



CHAPTER VI 
Intersection of Surfaces 



133. General principles 99 

134. Character of Auxiliary Cutting Surfaces . 99 

135. To determine the curve of intersection be- 

tween a cone and cylinder with axes oblique 

to the coordinate plane .... 100 

136. Order and Choice of Cutting Planes . .103 

137. To determine if there be one or two curves 

of intersection 103 

138. To determine the visible portions of the 

curve 104 

139. To determine the curve of intersection be- 

tw^een two cylinders, the axes of which are 
oblique to the coordinate planes . .104 



140. To determine the curve of intersection be- 

tween two cones, the axes of which are 
oblique to the coordinate planes . . 104 

141. To determine the curve of intersection be- 

tween an ellipsoid and an oblique cylinder 105 

142. To determine the curve of intersection be- 

tween a torus and a cylinder, the axes of 
which are perpendicular to the horizontal 
coordinate plane 106 

143. To determine the curve of intersection be- 

tween an ellipsoid and a paraboloid, the 
axes of which intersect and are parallel to 
the vertical coordinate planes . . . 108 



CHAPTER VII 
Warped Surfaces 



144. Warped Surface. Classification . 

145. Having given three curvilinear directrices 

and a point on one of them, it is required 



109 



to determine the two projections of the 
element of the M'arped surface passing 
through the given point .... 



110 



TABLE OF CONTENTS 



146. Having given two curvilinear directrices and 

a plane director, to draw an element of 
the warped surface, Case 1 . . . 

147. Having given two curvilinear directrices and 

a plane director, to draw an element of 
the warped surface. Case 2 . . . 

148. Modifications of the types in Arts. 145 and 

146 

149. The Hyperbolic Paraboloid .... 

150. Through a given point on a directrix to draw 

an element of the hyperbolic paraboloid . 

151. Having given one projection of a point on an 

hyperbolic paraboloid, to determine the 
other projection, and to pass an element 
through the point 



Ill 



112 

112 
114 

116 



116 



AKT. 

1.52. 
153. 
154. 
15.5. 
156. 

157. 



159. 
160. 



Warped Helicoids . „ , . . 

Right helicoid 

General type of warped helicoids . 

Hyperboloid of Revolution of one Nappe 

Thi'ough any point of the surface to draw 
an element ....... 

The Generatrix may be governed by Three 
Rectilinear Directrices , . . 

The Generatrix may be governed by Two 
Curvilinear Directrices and a cone Diiector 

Tangent plane to an hyperboloid of revolu- 
tion ........ 

Through a right line to pass a plane tangent 
to any double-curved surface of revolution. 
General solution ... 



PAGE 

118 
118 
118 
118 



120 
121 
121 
121 

122 



161. Directions for solving the problems 



CHAPTER Vm 
Problems 
. 124 162. Problems 



125 



DESCRIPTIVE GEOMETRY 



DESCRIPTIVE OEOMETRY 



CHAPTER I 



DEFINITIONS AND FIRST PRINCIPLES 



I. Descriptive Geometry is the art of graph- 
ically solving problems involving three dimen- 
sions. By its use the form of an object may 
be graphically defined, and the character, 
relation, and dimensions of its lines and sur- 
faces determined. 

To the student it presents the most admir- 
able training in the use of the imagination, 
such as the engineer or architect is called upon 
to exercise for the development of new forms 
in structure and mechanism, and which must 
be mentally seen before being graphically ex- 
pressed. 

To the engineer and architect it supplies 
the principles for the solution of all problems 
relating to the practical representation of forms, 



as illustrated by the various types of working 
drawings which are used by artisans to exe- 
cute designs. 

It is the foundation for the understanding 
of the different systems of projection such as 
orthographic, oblique, and perspective. 

2. Projection. The representation of an 
object is made on one or more planes by a 
process known as projection, the picture, draw- 
ing, or projection of the object being deter- 
mined by the intersection of a system of lines 
with the plane. The lines are known as pro- 
jectors and are drawn from the object to the 
plane of projection, or picture plane. If these 
lines, or projectors, are perpendicular to the 
plane of projection, the system is known as 



DESCRIPTIVE GEOMETRY 



Orthographic Projection, and at least two pro- 
jections, or views, are required to fully repre- 
sent the object. Fig. 1 represents two views 
of a box by orthographic projection. This 
system is the one commonly employed for 
working drawings, and for the solution of 
problems in Descriptive Geometry. 

If the projectors are parallel to each other, 
and oblique to the plane of projection, the 
system is known as Oblique Projection.* Fig. 2 
represents the application of this method to the 
illustration of the object shown in Fig. 1. This 
system is used for the purpose of producing a 
more pictorial effect, but one which is easily 
executed and susceptible of measurement. 

If the projectors converge to a point on the 
opposite side of the plane of projection, the 
system is known as Perspective. Fig. 3 is an 
application of this method to the representa- 
tion of the object previously illustrated. This 
system is used to produce the pictorial effect 

*For a treatise on Oblique Projection, and that branch 
of Orthographic Projection known as Isometric Projection, 
see "Elements of Mechanical Drawing" of this series. 



obtained by the camera, and is chiefly em- 
ployed by architects for the representation of 
buildings as they will appear to the eye of an 
observer. 

3. Coordinate Planes. The planes upon 
which the representations are made are called 
coordinate planes^ or planes of projection^ and 
are usually conceived to be perpendicular to 
each other and indefinite in extent. Fig. 4 
shows their relative positions, but for conven- 
ience of representation they are limited in 
extent. The plane designated by ^is called 
the horizontal coordinate plane, and the repre- 
sentation made thereon is known as the hori- 
zontal projection, plan, or top vieiu. The plane 
designated by F'is called the vertical coordinate 
plane, and the representation made thereon is 
known as the vertical projection, elevation, or 
front view. The plane designated by P is 
called tlie profile coordinate plane, and the 
representation made thereon is known as the 
p)rofile projection, side elevation, end or side 
view. The line of intersection between the V 
and ^planes is known as the ground line. 



ORTHOGRAPHIC PROJECTION 



4. Quadrants or Angles. The i^ortion of 
space lying in each of the four diedral angles 
formed by the vertical and horizontal coordi- 
nate planes is designated as follows : 

1st quadrant, above ^and before V. 
2nd quadrant, above ^and beliind V. 
3rd quadrant, below 5" and behind V. 
4th quadrant, below IT and before V. 

5. Orthographic Projection. The projection* 
of a point on any coordinate plane is obtained 
by letting fall a perpendicular from the point 
in space to the coordinate plane, its intersec- 
tion with that plane being the projection of 
the point. The perpendicular is called the 
projector or projecting line. In Fig. 4, a is the 
point in space and its projections are designated 
by the same letter with v^ A, or p written 
above and to the right, as a% signifying the 
vertical projection, a'\ the horizontal projec- 
tion, and a^, the profde projection. 

For convenience of representation, the verti- 
cal coordinate plane ^ is conceived as revolved 

*" Projection " used without a qualifying adjective 
always means orthographic projection. 




HORIZONTAL PROJECTION 



/ , 1 


^I~^ 


^ 


/ 
/ 




7 

J 


V 







/ 



VERTICAL PROJECTION 

Fig. I. 



OBLIQUE PROJECTION 
Fig. 2. 



PERSPECTIVE VIEW 
Fig. 3 









^Fh^^"" 




% i ^a 

\o GROUND LINE 



DESCRIPTIVE GEOMETRY 



about the ground line to coincide with the hori- 
zontal coordinate plane in such a way that the 
first and third quadrants will be opened to 
180°, and the second and fourth quadrants will 
be closed to 0°. In comparing Figs. 5, 6, and 
7 it will be observed that any point in the 
fourth quadrant, as point a, has, after the 
planes have been folded together, both pro- 
jections below the ground line; any point in 
the third quadrant, as point 5, has the vertical 
projection below, and the horizontal projection 
above, the ground line ; any point in the second 
quadrant, as point (?, has both projections above 
the ground line; and any point in the first 
quadrant, as point c?, has the vertical projection 
above, and the horizontal projection below, the 
ground line. Thus in Fig. 7 it is evident that 
the portion of the paper above the ground line 
represents not only that part of the horizontal 
coordinate plane wliich lies behind the vertical 
coordinate plane, but also that part of the ver- 
tical coordinate plane which lies above the 
horizontal coordinate plane. Likewise the 
paper below the ground line represents that 



portion of the horizontal coordinate plane ly- 
ing in front of the vertical coordinate plane, 
and also that portion of the vertical coordinate 
plane lying below the horizontal coordinate 
plane. 

The profile coordinate plane is commonly 
revolved about (r^-^Z as an axis until it coincides 
with the vertical coordinate plane, as in Figs. 
8 and 9; but it may be revolved about Gs L as 
an axis until it coincides with the horizontal 
coordinate plane, as in Figs. 10 and 11. It 
makes no difference as to the correct solution 
of the problem wliether the profile coordinate 
plane be revolved to the right or to the left, 
but care should be used to revolve it into such 
a position as will cause the least confusion with 
other lines of the problem. 

6. Notation. Points in space are designated 
by small letters, as a, 5, c^ etc., and their vertical, 
horizontal, and profile projections by the same 
letters with v, A, and p placed above and to the 
right, as 6»^, a^ a^. When revolved into one 
of the coordinate planes, the points will be 
designated by a\ 6', or a", J", etc. 



COORDINATE PLANES 




J G- 



^ ^/r 



Fig. 9. 



\ 




\ 


x^ ^^ 




\..'. 

^.a" 


V and H *cjh 






Fig. 6. 





c" 



•6" 



d" 



ORTHOGRAPHIC PROJECT/ON 

Fig. 7. 




cT 






1 


(/' 


r d" 



Fig. 10 



Fig. I 



6 



DESCRIPTIVE GEOMETRY 



A line in space is designated by two of its 
points, as ah, or by a capital letter, usually 
one of the first ten of the alphabet, as J., and 
its projections are designated by a'^h'^, a^h^, 
aPh^, or J.^, A^, A^\ When revolved into one 
of the coordinate planes, the lines will be desig- 
nated by a'h\ a"b", or A\ A", etc. The trace 
of a line, i.e. the point in which it pierces a 
coordinate plane, may be designated by V-tr, 
ITtr, or P-^r, according as the line pierces the 
vertical, horizontal, or profile coordinate plane. 

A plane in space is determined by three 
points not in the same straight line ; by a line 
and a point ; or by two parallel or intersecting 
lines. In projection a plane is usually desig- 
nated by its traces, i.e. the lines in which it 
pierces the coordinate planes. These traces 
are designated by the last letters of the alpha- 
bet beginning with M, as VM, HM, PM, accord- 
ing as the plane M intersects the vertical, 
horizontal, or profile coordinate plane. 

Abbreviations. The following abbreviations 
are used in connection with the figures and 
problems : 



Fi the vertical coordinate plane. 

H, the horizontal coordinate plane. 

P, the profile coordinate plane. 

(rX, ground line, the line of intersection 
between l^and H. 

VP^ the intersection between T^and P. 

HP, the intersection between H and P. 

IQ, 2Q, 3Q, 4Q, the first quadrant, second 
quadrant, etc. 

In general, an object in space is definitely 
located by two projections only, usually the 
vertical and the horizontal. Hereafter, only 
these two projections will be considered, unless 
otherwise indicated. 

The character of the lines employed is as 
follows : 

. Given and required lines. 

Invisible and projection lines. 

. . All lines not designated above. 

7. Points. Two projections are necessary 
to locate a point with reference to V and H. 
From Figs. 12 and 13 it is obvious that these 
two projections must always lie on a perpen- 
dicular to the ground line after V has been 



POINTS 



revolved to coincide with //. The distance 
from the point in space to H is equal to the 
distance from its vertical projection to the 
ground line, and the distance from the point 
in space to T^is equal to the distance from its 
horizontal projection to the ground line. A 
point on either coordinate plane is its own 
projection on that plane, and its other projec- 
tion is in the ground line, ;;s points c, c?, and g, 
Figs. 12 and 13. 

The points represented in Figs. 12 and 13 
are described as follows: 

Point «, in 5$, 5 units from T'and 8 units 

from H. 
Point 6, in 2Q, -^ units from V and T units 

from H. 
Point /, in 4Q, 2 units from V and 6 units 

from H. 
Point c, in J£^ between IQ and 4Q, 4: units 

from J\ 
Point (/, in 1% between IQ and 2Q^ 7 units 

from H. 
Point e, in GL. 



8. Lines. Since a line in space is deter- 
mined Ijy any two of its ^^oints, the projections 
of these two points determine the projections 
of the line, Figs. 14 and 15. A line may also 
be projected by passing planes through it 




b" 



•b 






Fig. 13. 

rrrrr 



SCALE 

TTfiT 



TTl 



8 



DESCRIPTIVE GEOMETRY 



perpendicular to the coordinate planes. The 
intersections of these planes with V^ H^ and P 
will determine respectively the vertical, hori- 
zontal, and profile projections of the line. 
Such auxiliary planes are known 'd.^ plane pro- 
jectors or projecting planes of the lines. In 
Fig. 14 the plane abh^'a'' is the vertical project- 
ing plane of the line ab\ the plane ahV'a^ is 
the horizontal projecting plane of the line ah^ 
and the plane ahh^a^ is the profile projecting 
plane of the line ah. 

9. A line parallel to a coordinate plane 
will have its projection on that plane parallel 
to the line in space, and its other projection 
will be parallel to the ground line. Line ^, 
Figs. 16 and 17, is a line parallel to V. A" is 
parallel to A in space, and A!" is parallel to GiL, 
Line B is parallel to ^, line Q is parallel to 
both Fand ^, and line I) is parallel to P. 

10. A line perpendicular to a coordinate 
plane will have for its projection on that plane, 
a point, and its other projection will be per- 
pendicular to the ground line. Line ^, Figs. 
18 and 19, is a line perpendicular to V. E"" is 



a point and E^ is perpendicular to GrL. Line 
jPis perpendicular to H, and line ^is perpen- 
dicular to P, K"^ and K'' being perpendicular 
to VP and HP^ respectively, and K^ being a 
point. 

11. A line lying in either coordinate plane 
is its own projection on that plane, and its 
other projection is in the ground line. In 
Figs. 20 and 21, line A lies in H and line B 
lies in V. 

12. A line parallel to one coordinate plane 
and oblique to the other has its projection on 
the plane to which it is parallel equal to the 
true length of the line in space, and the angle 
which this projection makes with tlie ground 
line is the true size of the angle which the line 
makes with the plane to which it is oblique. 
Line A, Figs. 16 and 17, is seen in its true 
length in its vertical projection, and it makes 
an angle of 30° with H. 

13. If two lines are parallel in space, their 
projections will be parallel. Figs. 22 and 23. 
Lines Cand D are parallel; therefore C^ and 
D'' are parallel, and 0^ and D^ are parallel. 



LINES 




Fig. 14 



Fig. 15. 



Fig. I 






B" 


Fig. 16. 










J^\ 


Q 




--/ 


A 


f 1 


\ 1 1 

1 ' ' 


1 


[ 


1 A' 1 


^\^>^ 


1 1 c^ 


. 

Q 





/ 


/ 
/ 
/ 
/ 






Fig. 17 


/ 




Fig. 20. 




a 

r 


K" 






\ \ 


\ 




1 1 1 
1 1 I 
1 1 1 

;^ 1 1 
1 z^'' ! 


Kl\ 







Fig. 19 



I p/. I 



I I 



Fig. 21 




10 



DESCRIPTIVE GEOMETRY 



14. If two lines intersect in space, they 
have one point in common; hence, the projec- 
tions of the lines will intersect each other in 
the projections of the j^oint, as at a^ and a'*, 
Fig§. 24 and 25. If the projections of the 
lines do not intersect on a common perpendicu- 
lar to the ground line, the lines in space do 
not intersect. Lines A and B^ Fig. 26, do 
not intersect. 

15. If a line intersects the ground line, as 
in Fig. 27, its projections will intersect the 
ground line in the same point. 

16. The traces of a line are the points in 
which the line pierces the coordinate planes. 
In Figs. 28 and 29, a is the horizontal trace, 
and h the vertical trace, of line C. The ver- 
tical projection of the horizontal trace is in 
the ground line, as is likewise the horizontal 
projection of the vertical trace. 

17. For convenience of expression it is cus- 
tomary to define the position of a line with 
respect to the coordinate planes by giving the 
quadrant in which it lies, together with its 



inclination with, and distance from, the coordi- 
nate planes. Line (7, Figs. 28 and 29, lies in 
the first quadrant, and if read from a toward 
^, inclines upward, backward, and toward the 
right. The vertical projection indicates that 
the inclination is upward; the horizontal pro- 
jection indicates that the inclination is, at the 
same time, backward; while either projection 
indicates that the inclination is to the right. 
The angle of inclination will be considered 
later. If the line be read from h toward a, the 
inclination would be downward, forward, and 
to the left. Either is correct.. Fig. 30 illus- 
ti'ates four other lines as follows : 

Line ab^ in 5§, inclined upward, forward, 

and to the right. 
Line ed^ in 2Q^ inclined downward, back- 
ward, and to the right. 
Line ef^ in 4Q^ parallel to H^ inclined for- 
ward and to the right. 
Line gk^ in 5§, parallel to P, and inclined 
upward and forward. 



LINES 



11 




Fig. 26 



12 



DESCRIPTIVE GEOMETRY 



1 8. Planes. The position of planes may be 
represented in projection as follows: 

1. By the projections of two intersecting or 
parallel lines. 

2. By the projections of a line and a point. 

3. By the projections of three points not in 
the same straight line. 

4. By the lines of intersection with the 
coordinate planes. (Traces.) 

All planes being indefinite in extent must 
intersect one or both of the coordinate planes. 
Such lines of intersection are called the traces 
of the planes. Fig. 31 illustrates the inter- 
sections of the planes N' with V and H, the 
vertical and horizontal traces being lettered 
FTVand ffiV, respectively. The orthographic 
representation is shown in Fig. 32, save that 
it is not always customary to continue the 
traces beyond the ground line, the horizontal 
trace being drawn on one side and the vertical 
trace on the other, as in Fig. 40. 

Since the vertical trace of a plane is a line 
lying on FJ it may also be lettered as the verti- 
cal projection of a line. Thus, in Figs. 31 



and 32, VJSf may be lettered A"^ and, according 
to Art. 11, Page 8, A^ must coincide with G-L. 
Likewise, HH is a line lying in IT and may 
be lettered 5\ w^iile B^ must lie in GrL. 

The following positions of planes are illus- 
trated by Figs. 33 to 48 inclusive: 

Perpendicular to 5" and parallel to FJ Figs. 

33 and 34. 
Perpendicular to T^and parallel to ff. Figs. 

35 and 36. 
Inclined to T^and H, but parallel to GL, 

Figs. 37 and 38. 
Inclined to ^and perpendicular to FJ Figs. 

39 and 40. 
Inclined to V'dnd perpendicular to ff^ Figs. 

41 and 42. 
Perpendicular to V and H, Figs. 43 and 44. 
Inclined to Fand IT^ but containing GL^ 

Figs. 45 and 46. 
Inclined to F, R, and GL, Figs. 47 and 48. 
The traces of parallel planes are parallel. 
19. From the foregoing illustrations it will 
be observed that the ground line is the hori- 
zontal projection of the vertical coordinate 



PLANES 



13 



plane, and that any point, line, or plane lying 
in Fwill have its entire horizontal projection 
in the ground line. Likewise, it will be ob- 
served that the ground line is the vertical 
projection of the horizontal coordinate plane, 
and that any point, line, or plane lying in H 
will have its entire vertical projection in the 
ground line. 



HN 




Fig. 31. 




Fig. 33. 



Fig. 34. 



VN 



Fig. 45 



VN 



Fig. 35. Fig. 36. 




Fig. 47. Fig. 48. 



CHAPTER II 



POINTS, LINES, AND PLANES 



20. Three distinct operations are required 
for the solution of problems in Descriptive 
Geometry. 

First, a statement of the Principles in- 
volved. 

Second, an outline of the Method to be 
observed, by the enumeration of the steps 
necessitated. 

Third, the graphic Construction of the 
problem. The first two operations are purely 
mental, and the last is the mechanical opera- 
tion of executing the drawing. 

21. To determine three projections of a line. 
Principle. The projections of two points 

of a Une determine the projections of the line. 
Method, 1. Determine the vertical, hori- 
zontal, and profile projections of two points of 
the line. 2. Connect the vertical projections 
of the points to obtain the vertical projection of 



the line ; connect the horizontal projections 
of the points to obtain the horizontal projec- 
tion of the line ; and connect the profile pro- 
jections of the points to obtain the profile 
projection of the line. 

Construction. Figs. 49 and 50. Let it 
be required to determine the projections of a 
line passing through point a, in i^, 4 units 
from F, 7 units from H^ and through point 5, 
in 4Q, 16 units from F", 9 units from H', point 
a to be 12 units to the left of point h. On 
any perpendicular to G-L lay off ^^ 9 units 
below GrL and ^^ 16 units below GL (Art. 7, 
page 6). On a second perpendicular, 12 units 
to the left of h"^ and 5^ lay off a^, 7 units above 
G-L and a^ 4 units below GL. Connect a^ 
and h^ to obtain the vertical projection of the 
line, and connect a^ and h^ to obtain the hori- 
zontal projection of the line. 



14 



PROJECTIONS OF A LINE 



15 



To obtain a^ and 5^, assume the position of 
the profile plane, shown in Fig. 49 by its inter- 
section with V and H as VF and HP. Re- 
volve P about VP as an axis to coincide with 
V. Then a^ will lie on a line drawn tlirouorh 

o 

a" parallel to 6rX, and at a distance from VP 
equal to that of a'' from CrL. This is obtained 
by projecting a^ to HP and rcA'olving HP 
about X as a center to coincide with CrL, and 
projecting perpendicularly to meet tlie parallel 
to CrL through a" at aP. Obtain h^ in like 
manner. Connect a^ and h^ to obtain the pro- 
file projection of the line. 

22. If point e in 2Q be one of the given 
points, e^ and e" having been determined, 
obtain e^ as follows: Project e^ to HP^ re- 



volve HP to (rZ, and project perpendicularly 
to meet a line drawn parallel to CrL through 
e*' at eP\ but after e^ has been projected to HP 
it is imperative that HP be revolved in the 
same direction that it Avas revolved when the 
profile projections of the other points were de- 
termined. If the horizontal projection of one 
point be revolved, then the horizontal projec- 
tions of all points must be revolved, and ver- 
tical projections must not be revolved. 

23. P may be revolved in either of the 
directions shown by Figs. 49 and 51. Fig. 51 
represents tlie line when P lias been revolved 
about HP as an axis to coincide with H. 
Here it will be observed that the vertical pro- 
jections, and only these, have been revolved. 





Fig. 49. 



4 8 12 16 20 2 



Fig. 50. 



Fig. 51 



16 



DESCRIPTIVE GEOMETRY 



24. To determine the traces of a line. 

Principle. The traces of a line are the 
points in which the line pierces the coordinate 
planes. The projections of these traces must, 
therefore, lie in the projections of the line, and 
one projection of each trace will lie in the 
ground line (Art. 7, page 6). 

Method. 1. To obtain the vertical trace 
of the line, continue the horizontal projection 
of the line until it intersects the ground line ; 
this will be the horizontal projection of the 
vertical trace, and its vertical projection will 
be perpendicularly above or below the ground 
line in the vertical projection of the given 
line. 2. To obtain the horizontal trace of 
the line, continue the vertical projection of the 
line until it intersects the ground line ; this 
will be the vertical projection of the horizon- 
tal trace, and its horizontal projection will 
be vertically above or below the ground 
line in the horizontal projection of the given 
line. 

Case 1. When the line is inclined to F", 
H, and P. 



Construction. Figs. 52 and 53. Let it 
be required to determine the vertical trace of 
line A. Continue A^ until it intersects GrL 
in c^ which is the horizontal projection of the 
vertical trace. Next project this point to J.^, 
as at c^ wliich is the vertical projection of the 
vertical trace. 

The horizontal trace is similarly determined 
thus : Continue A^ until it intersects GiL in 
d'"^ which is the vertical projection of the hori- 
zontal trace. Next project this point to J.^ 
as at d^^ which is the horizontal projection of 
the horizontal trace. 

To determine the profile trace, consider P, 
Fig. 52, to be represented by its intersections 
with I^and H^ as VP and ZTP, and to be re- 
volved to the right about VF as an axis until 
it coincides with V, Continue A^ until it 
intersects VP in/% which is the vertical pro- 
jection of the profile trace. Continue A^ until 
it intersects HP in /^ which is the horizontal 
projection of the profile trace, revolve to GrL^ 
using the intersection of HP with GrL as a 
center, and project to f^ by a line drawn 



TRACES OF A LINE 



17 



through/'' parallel to GrL. This is the profile 
projection of the profile trace. 

Fig. 53 is the oblique projection of the line, 
and clearly shows that the line passes from 
one quadrant to another at its vertical and 
horizontal traces, i.e. line A passes from IQ 
to 5 § at its vertical trace, c, and from i^ to 



4Q at its horizontal trace, d. 

Note. The vertical projection of the verti- 
cal trace of a line is often called the vertical 
trace of the line, since this trace and its ver- 
tical projection are coincident. Likewise the 
horizontal projection of the horizontal trace is 
called the horizontal trace of the line. 





18 



DESCRIPTIVE GEOMETRY 



25. Case 2. When the line is inclined to 
V and H and is parallel to P. 

CoNSTEUCTiON. Fig. 5.1:. If the given line 
is parallel to P, its vertical and horizontal 
traces cannot be determined by the above 
method, since the projections of the line are 
perpendicular to the ground line; hence, a 
profile projection of the line is necessary. 




T^' Fig. 54. . 

Let it be required to determine the vertical 
and horizontal traces of the line ab. P is 
assumed at will and is indicated by HP and 
VP. Determine a^-h^^ the profile projection of 
the line (Art. 21, page 14), by revolving P 
about VP as an axis until it coincides with V. 



That portion of P which before revolution 
was in 3Q now falls below GiL and to the 
right of VP \ that portion which was in 4Q 
falls below GrL and to the left of VP\ that 
portion which was in i ^ falls above CrL and 
to the left of HP. Thus the profile projec- 
tion of the line, when continued, indicates 
that the line passes through 3Q into 2§ at 
point d, dp being the profile projection of the 
horizontal trace. Likewise the line passes 
from SQ into 4^ at point c, c^ being the profile 
projection of the vertical trace. Counter- 
revolve P to its original position and obtain 
d^ and c?^, which are the horizontal and vertical 
projections of the horizontal trace, and c^ and 
c^ which are the vertical and horizontal pro- 
jections of the vertical trace. Since line ab is 
parallel to P, it has no profile trace. 

26. To determine the projections of a line 
when its traces are given (Art. 21, page 14). 

Method. 1. Determine the horizontal 
projection of the vertical trace, and the ver- 
tical projection of the horizontal trace. 
2. Connect the horizontal trace with the hori- 



LINES LYING IN PLANES 



19 



zontal projection of the vertical trace to obtain 
the horizontal projection of the line. Connect 
the vertical trace with the vertical projection 
of the horizontal trace to obtain the vertical 
projection of the line. 

27. Conditions governing lines lying in a 
plane. Since the traces of a plane are lines of 
the plane, they must intersect all other lines of 
the plane, and conversely, all lines of a plane 
must have their traces in the traces of the 
plane. See lines A and B, Figs. 5b and 56. 



28. If a line is parallel to IT, it intersects 
H at infinity; therefore, its horizontal trace is 
at infinity and its horizontal projection is 
parallel to the horizontal trace of the plane in 
which it lies, its vertical projection being 
parallel to the ground line. See line (7, Figs. 
57 and 58. Likewise if a line is parallel to V^ 
its vertical projection is parallel to the vertical 
trace of the plane in which it lies, and its 
horizontal projection is parallel to the ground 
line. 




Fig\55 



Fig. 58. 



20 



DESCRIPTIVE GEOMETRY 



29. If the traces of any plane be drawn 
through the traces of a Ime, the plane must 
contain the line ; therefore, an infinite number 
of planes may be passed through any line. In 
Fig. 59, planes iV, R, T, and S all contain line 
E. 

30. To pass a plane through two intersect- 
ing or parallel lines. 

PtiiNCiPLE. The traces of the plane must 
contain the traces of the lines (Art. 27, 
page 19). 

Case 1. Method. 1. Determine the traces 
of the given lines. 2. Connect the two hori- 
zontal traces of the lines to obtain the horizon- 
tal trace of the plane, and connect the two 
vertical traces of the lines to obtain the 
vertical trace of the plane. 

Construction. Fig. 60. A and B are the 
given intersecting lines. Determine their 
horizontal traces c^ and d}^ and their vertical 
traces e'' and /^ (Art. 21, page 16). Con- 
nect the horizontal traces to determine HT 
and the vertical traces to determine VT. 
Since the traces of the plane must meet in GL^ 



only three traces of the lines are necessary. 

Check. Both traces of the plane must 
intersect the ground line in the same point. 

Note. The vertical projection of the ver- 
tical trace of a plane will always be spoken of 
as the vertical trace of the plane, but it must 
constantly be borne in mind that tlie vertical 
trace of a plane is a line lying on V and that 
its horizontal projection is in the ground line. 
Likewise the horizontal projection of the hori- 
zontal trace of a plane will be spoken of as the 
horizontal trace of that plane, but, as before, 
the horizontal trace is a line lying in H and its 
vertical projection is in the ground line. 

31. Case 2. Method. If the traces of 
the given lines cannot readily be found, new 
lines intersecting the given lines may be 
assumed, which, passing through two points 
of the plane, lie in it, and their traces, there- 
fore, are points in the traces of the required 
plane. 

Construction. Fig. 61. The two given 
intersecting lines are A and B^ the traces of 
which cannot be found within the limits of 



THE PLANE OF LINES 



21 



the drawing. Line C is an assumed line join- 
ing point e of line A and point/ of line B^ the 
traces of which are easily located at g and h. 
Line i> is a second similar line. HT^ connect- 
ing the horizontal traces of lines C and Z>, is 
the required horizontal trace of the plane of 
lines A and B. Likewise YT, connecting the 
vertical traces of lines C and i), is the required 
vertical trace of the plane of lines A and B. 
32. Case 3. Method. If one of the given 



lines is parallel to the ground line, the required 
plane will be parallel to the ground line, and, 
therefore (Figs. 37 and 38, page 13), the 
traces of the plane Avill be parallel to the 
ground line. This problem may be solved by 
Case 2, or by the following method: Determine 
the profile trace of the required plane by ob- 
taining the profile traces of the given lines. 
Having found the profile trace of the plane, de- 
termine the horizontal and vertical traces. 




Fig. 59 




Fie-. 60 



22 



DESCRIPTIVE GEOMETRY 



Construction. Fig. 62. Let A and B be 
the given lines parallel to the ground line. 
Assume P and draw JIP and FP. Continue 
the horizontal and vertical projections of lines 
A and B to intersect KI* and FP, respectively. 
The horizontal and vertical projections of the 
profile trace of line A will lie at c^ and c^ and 
the profile trace at c^. Similarly determine c?^, 
the profile trace of line B, Through c'p and 
d^ draw PN^^ the profile trace of the required 
plane. U^ will be the profile projection of the 
horizontal trace of the required plane, and & 
the profile projection of the vertical trace. By 
counter-revolution obtain ^iVand FTV. 

33. To pass a plane through a line and a 
point. 

Method. Connect the given point with 
any assumed point of the line and proceed as 
in Art. 30, page 20. 

34. To pass a plane through three points 
not in the same straight line. 

Method. Connect the three points by aux- 
iliary lines and proceed as in Art. 30, page 20. 
In Fig. 63, a, J, and c are the given points. 



35. Given one projection of a line lying on 
a plane, to determine the other projection. 

Principle. The traces of the line must 
lie in the traces of the plane (Art. 27, page 
19). 

Method. Determine the traces of the line 
and from them determine the unknown pro- 
jection of the line. 

Construction. Fig. 64. LetffiVandFiV 
be the traces of the given plane, and A^ one 
projection of line A lying in iV. Continue A^ 
to meet IIJV in a\ the horizontal trace of line 
A; a^ is in GL (Art. 24, page 16). Con- 
tinue A^ to meet GL in 5\ the horizontal pro- 
jection of the vertical trace; 6^, the vertical 
trace, is in FiV. Connect a" and h^ to obtain 
^^ the required vertical projection of line A. 

If A" had been the given projection, A^ 
would have been similarly determined. 

If the given projection, B^, Fig. 65, be par- 
allel to (7X, B" will be parallel to HS\ for, if 
a line lying in an inclined plane has one pro- 
jection parallel to the ground line, the other pro- 
jection is parallel to the trace of the plane; and 



PROJECTION OF LINES IN PLANES 



23 



conversely, if a line lying in an inclined plane 
has one projection p>araUel to the trace of the 
plane^ the other projection is parallel to the 
ground line (Art. 28, page 19). Therefore, 
to determine B^ continue B^ to meet VS in 
c^ the vertical trace of the line. Its horizon- 
tal projection, c^, will be in GrL, and B'' will 
pass through c'* parallel to MS. 





f" HN 


a: 
A'' 


c" Xc" 


b' 


1 \ -o 

d'^ 1 "" d" 




1 \ Kf" 


A" c" 


I ; 1 
> 


B" d" 






_--^ VN 







62. 




24 



DESCRIPTIVE GEOMETRY 



36. Given one projection of a point lying on 
a plane, to determine the other projection. 

Pkinciple. The required projection of 
the point will lie on the projection of any line 
of the plane passed through the point. 

Method. 1. Through the given projec- 
tion of the point draw the projection of any 
line lying on the plane. 2. Determine the 
other projection of the line (Art. 35, page 
22). 3. The required projection of the point 
will lie on this projection of the line. 

Construction. Figs. 66 and 67. Let a^ 
be the given projection of point a on plane iV. 
Through a'* draw B^^ the horizontal projection 
of any line of pLme iV passing through point 
a. Determine B'' (Art. 35, page 22). Then 
a" lies at the intersection of B^ and a per- 
pendicular to GL through a^. The same 
result is obtained by using line lying in 
plane iV and parallel to V, or using line I) 
lying in plane iVand parallel to IT. 

If the vertical projection of the point be 
given, the horizontal projection is determined 
in a similar manner. 



In general, solve the problem by the use of 
an auxiliary line parallel to V or IT. 

37. To locate a point on a given plane at a 
given distance from the coordinate planes. 

Principle. The required point lies at the 
intersection of two lines of the given plane; 
one line is parallel to, and at the given dis- 
tance from, F, and the other line is parallel to, 
and at the given distance from, H. 

Method. 1. Draw a line of the plane 
parallel to, and at the required distance from, 
one of the coordinate planes (Fig. 65^ page 
23). 2. Determine a point on this line at 
the required distance from the other coordi- 
nate plane. 

Construction. Fig. 68. Let it be re- 
quired to locate point b on plane B, at x 
distance from B[ and y distance from V. 
Draw line A in plane i^, parallel to, 
and at ?/ distance from, V. A'^ will be 
parallel to, and at y distance from, G-L^ and 
A"^ will be parallel to VB (Art. 35, page 22). 
On A^ determine h'" at x distance from (7Z, 
and project to A!' to determine h^. Or, point 



TO REVOLVE A POINT 



25 




h may be located by passing line C in plane R^ 
parallel to, and at x distance from, H. 

If the plane be parallel to GiL^ use the fol- 
lowing method: 1. Draw on the given plane 
any line oblique to V and H. 2. Locate the 
required point thereon at the required distance 
from I^and H. 

38. To revolve a point into either coordi- 
nate plane. 

Principle. The axis about which the point 
revolves must lie in the j)lane into which the 
point is to be revolved. The revolving point 
will describe a circle whose plane is perpen- 
dicular to the axis, and whose center is in the 
axis. The intersection of this circle and the 
coordinate plane is the required revolved posi- 
tion of the point. 

Method. 1. Through that projection of 
the given point on the plane in which the axis 
lies draw a line perpendicular to the axis. 
2. On this perpendicular lay off a point hav- 
ing its distance from the axis equal to the 
hypotenuse of a right triangle, one leg of which 
is the distance from the projection of the point 



26 



DESCRIPTIVE GEOMETRY 



to the axis, and the other leg of which is equal 
to the distance from the second projection of 
the point to the ground line. 

Let a., Fig. 69, be the point in space to be 
revolved into H about D B^ lying in H as an 
axis. If a point be revolved about an axis, 
its locus will be in a plane perpendicular to 
the axis. X is this plane, and HX^ its hori- 



zontal trace, is perpendicular to B^. Point a^ 
in revolving, will describe a circle with ah as 
a radius. The points a' and a", in which this 
circle pierces H, are the required revolved 
positions of the point a. Since angle aa^h is 
a right angle, ha is equal to the hypotenuse of 
a right triangle, one leg of which is «^6, the 
distance from a^ to the axis, and the other leg 




Fig. 71 




Fig. 72 



TRUE LENGTH OF A LINE 



27 



is a^a^ or the distance from a^ to GL. 

Construction. Fig. 70. The point a is 
represented by its two projections, a" and a^. 
Through a!^ draw HX perpendicular to D^. 
The revolved position, a! or d\ will lie in HX 
at a distance from D^ equal to the hypotenuse 
of a right triangle, one leg of which is the dis- 
tance from a^ to the axis i>^ and the other leg 
is the distance from ay to GL. 

If the axis lies in Fi the revolved position 
of the point will lie in a line passing through 
the vertical projection of the point perpen- 
dicular to the axis and at a distance from this 
axis equal to the hypotenuse of a right tri- 
angle, one leg of which is the distance from 
the vertical projection of the point to the axis, 
while the other leg is the distance from the 
horizontal projection of the point to the ground 
line. 

39. To determine the true length of a line. 

Principle. A line is seen in its true length 
on that coordinate plane to which it is parallel, 
or in which it lies (Figs. 16 to 21, page 9). 

Method. Kevolve the line parallel to, or 



into either coordinate plane, at which time one 
of its projections wdll be parallel to the ground 
line and the other projection will measure the 
true length of the line in space. 

Case 1. When the line is revolved paral- 
lel to a coordinate plane. 

Construction. Fig. 71. Let ah be a line 
in the first quadrant inclined to both Fand 
H, Neither projection will equal the true 
length of the line, in space. Revolve ah about 
the projecting line aa!^ as an axis until it is 
parallel to F. Point a will not move ; hence, 
its projections, a^ and a^ will remain station- 
ary. Point h will revolve in a plane parallel 
to H^ to h^ ; hence, h^ will describe the arc h^h^^ 
and a^h^ will be parallel to GrL, Then 6^ will 
move parallel to GL to its new position 5^^, and 
a'^'hl^ the new vertical projection, will equal the 
true length of line ah. 

Fig. 72 is the orthographic projection of the 
problem. It is of interest to note that the 
angle which this true length makes with GrL 
is the true size of the angle which the line ii;i 
space makes with H. 



28 



DESCRIPTIVE GEOMETRY 



Figs. 73 and 74 represent the line ah re- 
volved parallel to H^ thus obtaining the same 
result as to the length of line. Here the 
angle which the true length of the line makes 
with CiL is the true size of the angle which 
the line in space makes with V. 

40. Case 2. When the line is revolved 
into a coordinate plane. 

Construction. Figs. 75 and 76 represent 
the line ah of the previous figures revolved 
into H about its horizontal projection a%^' as 
an axis. Point a revolves in a plane perpen- 
dicular to the axis a%^' (Art. 38, page 25); 
therefore, its revolved position, a\ will lie in a 
line perpendicular to a%^ and at a distance 
from a^ equal to aa^ or the distance from a'" to 
CrL.^ Similarly h' is located. 

41. From Fig. 76 it will be seen that if the 
revolved position of the line, a'h\ be continuec^ 
it will pass through the point where ah, con- 
tinued, intersects its own projection, which is 

* This does not contradict Art. 38, page 26, in that one 
leg of the triangle is equal to zero. 



the horizontal trace of the line, and since it is 
in the axis, it does not move in the revolution. 
Thus, in every case where a line is revolved 
into one of the coordinate planes, its revolved 
position will pass through the trace of the line. 
Figs. 77 and 78 illustrate a line ed piercing V 
in the point e. In order to determine its true 
length it has been revolved into ]^ about c^'d" 
as an axis. Since the point e lies in the axis 
and does not move, point c revolves in one 
direction while point d must revolve in the 
opposite direction, thus causing the revolved 
position to pass through the vertical trace of 
the line. 

Again it is of interest to note that when a 
line is revolved into a coordinate plane to de- 
termine its true length, the angle which the 
revolved position makes with the axis is the 
true size of the angle which the line in space 
makes with the coordinate plane into which it 
has been revolved. Thus in Figs. 77 and 78, 
angle c'e^'c" is the true size of the angle which 
line cd makes with V. 



TRUE LENGTH OF A LINE 



29 





Fig. 76. 



Fig. 77. 




Fig. 71 



30 



DESCRIPTIVE GEOMETRY 



42. Given a point lying on a plane, to 
determine its position when the plane shall 
have been revolved about either of its traces 
as an axis to coincide with a coordinate plane. 

Principle. This is identical with the prin- 
ciple of Art. 38, page 25, since either trace of 
the plane is an axis lying in a coordinate plane, 
one projection of which is the line itself, and 
the other projection of which is in the ground 
line. 

Method. See Art. 38, page 25. 

Construction. Fig. 79. ffiVandFiVare 
the traces of the given plane and 5'' and 5% the 
projections of the point. If the plane with 
point h thereon be revolved into H about HN 
as an axis, the point will move in a plane per- 
pendicular to ffiV, and will lie somewhere in 
h%'. Its position in this line must be deter- 
mined by finding the true distance of the point 
from HN^. This distance will equal the hy- 
potenuse of a right triangle of which h^d^ 
the horizontal projection of the hypotenuse, 
is one leg, and h^c^ the distance of the point 
from H^ is the other leg. 



/ 
/ 
/ 


a6' 

\ ^ 


\ 


« 1 


^/^ 


1 


^^~^^ 


^£:--v' 


Fig. 79 





By laying off V'e 




ANGLE BETWEEN LINES 



31 



equal to ch'"^ and perpendicular to h^d^ the length 
required, de^ is determined, and when laid off 
on dh^ from c?, will locate the revolved position 
of point h. 

If it were required to revolve the point into 
"F about FiVas an axis, it would lie at h" in a 
perpendicular to VN through h^. The dis- 
tance kh" will equal Icf^ the hypotenuse ■ of a 
right triangle, one leg of which is ^5% and the 
other leg of which is equal to ch^. 

43. The revolved position of a line lying 
in a plane is determined by finding the re- 
volved position of two points of the line. If 
the line of the plane is parallel to the trace 
which is used as an axis, its revolved position 
will also be parallel to the trace. 

If a line has its trace in the axis, this trace 
will not move during the revolution; there- 
fore, it will be necessary to determine but one 
other point in the revolved position. 

Fig. 80 represents the lines C and D of the 
plane R revolved into ^ about HR as an axis. 
The revolved position of point a is determined 



at a\ through which C is drawn parallel to 
HR (since line C of the plane is parallel to 
HR^, and D' through e'', the horizontal trace 
of line D. 

44. To determine the angle between two 
intersecting lines. 

Principle. The angle between intersect- 
ing lines maybe measured when the plane of the 
lines is revolved to coincide with one of the 
coordinate planes. 

Method. 1. Pass a plane through the two 
intersecting lines and determine its traces 
(Art. 30, page 20). 2. Revolve the plane 
with the lines thereon, about either of its 
traces as an axis, until it coincides with a co- 
ordinate plane (Art. 43, page 31). As it is 
necessary to determine the revolved position 
of but one point in each line, let the point be 
one common to both lines, their point of inter- 
section, and, therefore, the vertex of the angle 
between them. 3. The angle between the re- 
volved position of the lines is the required 
angle. 



32 



DESCRIPTIVE GEOMETRY 



45. To draw the projections of any polygon 
having a definite shape and size and occupying 
a definite position upon a given plane. 

Principle. The polygon will appear in its 
true size and shape, and in its true position on 
the plane, when the plane has been revolved 
about one of its traces as an axis to coincide 
with a coordinate plane. 

Method. 1. Revolve the given plane into 
one of the coordinate planes about its trace as 
an axis. 2. Construct the revolved position of 
the polygon in its true size and shape, and 
occupying its correct position on the plane. 
3. Counter-revolve the plane, with the poly- 
gon thereon, to its original position, thus 
obtaining the projections of the polygon. 

Construction. Fig. 81. Let it be re- 
quired to determine the projections of a regu- 
lar pentagon on an oblique plane iVJ the center 
of the pentagon to be at a distance x from H^ 
and y from F^ and one side of the pentagon to 
be parallel to FJ and of a length equal to z. 

Determine the projections of the center as 
at 0"" and 0^ (Art. 37, page 24). Revolve 



plane N about one of its traces as an axis 
until it coincides with one of the coordinate 
planes (in this case about VN until it coin- 
cides with V')^ 0' being the revolved position 
of the center (Art. 42, page 30). About 
0' as a center draw the pentagon in its true 
size and shape, having one side parallel to VJV, 
and of a length equal to z. Counter-revolve 
the plane to obtain the projections of the 
pentagon. 

46. Counter-revolution. The counter- 
revolution may be accomplished in several 
ways, of which three are here shown. 

Construction 1. Through 0', Fig. 81, 
draw any line 0' to intersect VJV in n^, and 
connect n'" with 0'" to obtain 0^. Through d' 
draw D' parallel to C to intersect VJ}^ in m'^\ 
from which point draw D^ parallel to C". 
From d' project perpendicularly to F7V to in- 
tersect D^ in c?% the vertical projection of one 
point in the required vertical projection of the 
pentagon. Similarly determine the vertical 
projections of the other points by drawing 
lines parallel to O. 



PROJECTIONS OF A POLYGON 



33 



47. Construction 2. A sometimes shorter 
method of counter-revolution, to obtain the 
vertical projection of the pentagon, is as 
follows. Fig. 82. Assume the revolved 
position of the pentagon to have been drawn 
as described above. From any point of the 
pentagon, as c\ draw a line through the center 
o\ and continue it to meet the axis of revolu- 
tion in m". Then m'o" is the vertical projec- 
tion of this line after counter-revolution (Art. 
43, page 31), and c" will lie on m'o'' at its inter- 




section with a perpendicular to VN from c' 
(Art. 42, page 30). Produce c'V to VN, thus 
determining the line c''5", and, therefore, If. 
Since V a^ is parallel to YN, a^' may be found 
by drawing a parallel to VJV from b^ to meet 
the perpendicular from a' (Art. 35, page 
22). Likewise d" is determined, since e'd' is 
parallel to VN. Next continue e'd' to the 
axis at 7^"; draw n''d" to meet o'd"" produced at 
e'". The horizontal projection of the pentagon 
is best determined by Art. 35, page 22. 




Fig. 82. 



34 



DESCRIPTIVE GEOMETRY 



48. Construction 3. Fig. 83. Let it be 
assumed that tlie plane iV has been revolved 
into V about FTV as an axis, and that the 
revolved position of UN' has been determined, 
as mSf', by revolving any point g of the hori- 
zontal trace of the plane about VJSf as an axis 
(Art. 43, page 31). The angle between RN^ 
and VJV is the true size of the angle between 
the traces of the plane ; also the area between 
RN ' and FiV is the true size of that portion 
of plane JV lying between its traces. Next let 
it be assumed that the polygon has been 
drawn in its revolved position, and occupying 
its correct location with respect to FiV and 
ffiV, e' and d^ being two of its vertices. To 
counter-revolve, continue e'd' to intersect RN' 
in k', and VN in ri". Since A; is a point in the 
horizontal trace of the plane, its vertical pro- 
jection will lie in GrL. Then P will lie in 
CrL at the intersection of a perpendicular to 
FiV from k', and k^ is in RK Also the inter- 
section of e'd' with the axis VW is n'^\ and its 
horizontal projection will lie in GL at n^. 
Draw rfk'".) producing it to intersect perpen- 



diculars to VJY from d' and e\ in d'" and e". 
Draw n^k^\ producing it to intersect perpen- 
diculars to GL from d'" and e'"^ at d^ and e^. 
Similarly determine the projections of the 
other points of the pentagon. 

A fourth construction for counter-revolu- 
tion is explained in Art. 91, page 63. 




Fig. 83 



INTERSECTION BETWEEN PLANES 



35 



49. To determine the projections of the line 
of intersection between two planes. 

PRINCIPLE. The line of intersection be- 
tween two planes is common to each plane; 
therefore, the traces of this line must lie in 
the traces of each plane. Hence, the point of 
intersection of the vertical traces of the planes 
is the vertical trace of the required line of 
intersection, and the point of intersection of 
the horizontal traces of the planes is the hori- 
zontal trace of the required line of intersec- 
tion between the planes. 

There may be three cases as follows : 

Case 1. When no auxiliary plane is re- 
quired. 

Case 2. When an auxiliary plane parallel 
to V OT ^is used. 

Case 3. When an auxiliary plane parallel 
to P is required. 

50. Case 1. When like traces of the 
given planes may be made to intersect, no 
auxiliary plane is required for the solution of 
the problem. 

Method. 1. Determine the points of in- 



tersection of like traces of the planes, which 
points are the traces of the line of intersection 
between the planes. 2. Draw the projections 
of the line (Art. 26, page 18). 

Construction. Figs. 84 and 85 illustrate 
the principle and need no explanation, line ch 
being the line of intersection between the 
given planes iVand S. 




Fig. 84. 



Fig. 85. 



36 



DESCRIPTIVE GEOMETRY 



Figs. 86 and 87 illustrate a special condi- 
tion of Case 1, in which one of the planes is 
parallel to a coordinate plane. Plane iV is 
inclined to both Fand H^ and plane R is par- 
allel to H. Since plane R is perpendicular to 
Fi its vertical trace will be the vertical pro- 
jection, ^y^ of the required line of intersec- 
tion between planes N and R. As e" is the 
vertical trace of line ^/, the horizontal projec- 
tion of this line will be ^y^ which is parallel 
to HN (Art. 35, page 22). 

51. Case 2. When one or both pairs of 
like traces of the given planes do not intersect 
within the limits of the drawing, and are not 
parallel, or when all the traces meet the 
ground line in the same point, auxiliary cut- 
ting planes parallel to either For ^ may be 
used for the solution of the problem. 

Method. 1. Pass an auxiliary cutting 
plane parallel to V or H^ and determine the 
lines of intersection between the auxiliary 
plane and each of the given planes. Their 
point of intersection will be common to the 
given planes, and, therefore, a point in the re- 



quired line. 2. Determine a second point by 
passing another auxiliary plane parallel to V 
or H^ and the required line will be determined. 

Construction. Fig. 88 represents two 
planes, T and Jf, with their horizontal traces 
intersecting, bat their vertical traces inter- 
secting beyond the limits of the drawing. 

By Case 1, point d is one point in the line 
of intersection between the planes. To obtain 
a second point an auxiliary plane -X" has been 
passed parallel to V. Then HX is parallel to 
GrL^ and C^ the horizontal projection of the 
line of intersection between planes X and M, 
lies in HX, while (7^ is parallel to VM (Case 
1). Likewise B^, the horizontal projection of 
the line of intersection between planes X and 
r, lies in HX, while B' is parallel to VT. 
Then point a, the point of intersection between 
lines B and C, is a point in the line of inter- 
section between planes il[f and T, since point 
a lies in line C of plane M, and in line B of 
plane T. Line ad is, therefore, the required 
line of intersection between the two given 
planes. 



INTERSECTION BETWEEN PLANES 



37 




Fig. 87. 




If neither the vertical nor horizontal traces 
of the given planes intersect within the limits 
of the drawing, it will be necessary to use two 
auxiliary cutting planes, both of which may be 
parallel to F", both parallel to H^ or one parallel 
to F"and one parallel to H. 

52. If two traces of the given planes are 
parallel, the line of intersection between them 
will be parallel to a coordinate plane, and will 
have one projection parallel to the ground line 
and the other projection parallel to the parallel 
traces of the given planes. 

53. Fig. 89 represents a condition when all 
four traces of the^ given planes intersect GL 
in the same point, neither plane containing 
GL. This point of intersection of the traces, 
5, is one point in the required line of inter- 
section between the planes (Case 1). A 
second point, a, is obtained by passing the 
auxiliary cutting plane X parallel to JT, inter- 
secting plane N in line Q and plane S in line 
B. Point a, the point of intersection of these 
lines, is a second point in the required line of 
intersection, ah^ between the given planes N 
and S (Case 2). 



38 



DESCRIPTIVE GEOMETRY 



54. Case 3. When both intersecting 
planes are parallel to the ground line, or 
when one of the intersecting planes contains 
the ground line. 

Method. 1. Pass an auxiliary cutting plane 
parallel to P. 2. Determine its line of inter- 
section with each of the given planes. 
3. The point of intersection of these two lines 
is one point in the required line of intersection 
between the two given planes. It is not nec- 
essary to determine a second point, for, when 
both given planes are parallel to the ground 
line, their line of intersection will be parallel 
to the ground line. 

When one plane contains the ground line, 
one point in the line of intersection is the point 
of intersection of all the traces. (Figs. 91, 92.) 

Construction. In Fig. 90 the two planes 
iVand /S' are parallel to CrL. Pass the auxil- 
iary profile plane P intersecting plane N in 
the line whose profile projection is D^, and the 
plane S in the line whose profile projection is 
B^. A^, the intersection of these two lines, is 
the profile projection of one point in the line 



of intersection between planes N and S ; in 
fact A^ is the profile projection of the re- 
quired line, and A^ and A^ are the required 
projections. 

55. In Fig. 91 the plane 8 contains GrL and, 
therefore, it cannot be definitely located with- 
out its profile trace, or its angle with a coordi- 
nate plane, and the quadrants through which 
it passes. Plane iVis inclined to V,II^ and P, 
and plane S contains (xZ, passing through 1 Q 
and 3Q at an angle with V. As in the pre- 
vious example the profile auxiliary plane is 
required and PN^ the profile trace of iV, is 
determined as before. PS^ the profile trace 
of S, is next drawn through 1 Q and 3 Q and 
making an angle 6 with VP. Then c?^, the 
intersection of PiVand PS^ is the profile projec- 
tion of one point in the required line of inter- 
section between planes N and S^ and d^ and 
d^ are the required projections of this point. 
The horizontal and vertical traces of this line 
of intersection are at 5, for it is here that VN 
and VS intersect, and also where ffiVand HS 
intersect. The projections of two points of 



INTERSECTION BETWEEN PLANES 



39 



the line having now been determined, the pro- 
jections of the line, dh, may be drawn. 

Case 3 is applicable to all forms of this prob- 
lem, and Fig. 92 represents the planes iV^and 
8 of Fig. 89 with their line of intersection, ah, 
determined by this method. 





HS 




a: 


V 

\ 


HN 




^^/l'' 


a' 




1 \ ^^^^^ 




1 1 
/ / 

y 1 


1 
A' / 




/ 


/ 
VS / 


^ 




/ 

/ 
/ 

/ 
y 
y 

VN 




Fig. 90. 






40 



DESCRIPTIVE GEOMETRY 



56. Revolution, Quadrants, and Counter-rev- 
olution. Fig. 93. Let it be required to pass 
a plane S through the first and third quadrants, 
making an angle 6 with P^and intersecting the 
oblique plane iV in the line A. The problem is 
solved by Art. 54, page 38, but this question 
may arise: In what direction shall Pas' be drawn 
and with what line shall it make the given 
angle 6? Conceive the auxiliary profile plane 
P to be revolved about VP as an axis until it 
coincides with F, that portion of P which was 
in front of V to be revolved to the right. 
Then the line VP HP represents the line of 
intersection between F^and P; (77> represents 
the line of intersection between H and P; and 
PN^ the line of intersection between iVand P. 
That portion of the paper lying above GrL and 
to the right of VP will represent that portion 
of P which, before revolution, was in IQ; 
above GL and to the left of VP^ in 2Q; below 
GrL and to the left of VP^ in 3Q; and below 
GL and to the right of FP, in 4Q. Since iS 
is to pass through IQ and 3Q, making an angle 
6 with F", the line PS, in which >S' intersects 



P, should be drawn on that portion of P which 
is in IQ and 3Q, and should pass through the 
point of intersection of VP and (rX, making 
the angle 6 with VP. PS'dnd PiV intersect in 
cZ^, the profile projection of one point in the 
required line of intersection. In counter- 
revolution, that is, revolving P back into its 
former position perpendicular to both Fand ZT, 
rotation will take place in a direction opposite 
to that of the first revolution, and d^ and d^ 
will be as indicated. 

Fig. 94 represents the same problem when 
the auxiliary profile plane P has been revolved 
in the opposite direction to coincide with V 

Figs. 95 and 96 are examples of the same 
problem when the auxiliary profile plane P 
has been revolved about JIP as an axis until it 
coincides with S. Tlien CrL will represent 
the revolved position of the line of intersection 
between V and P, and the line VP JIP will 
represent the revolved position of the line of 
intersection between ^and P. Pas' will then 
make its angle 6 with GrL, and its direction 
will be governed by the rotation assumed. 



COUNTER-REVOLUTION 



41 




Fig. 93. 






3Q 


<^ 




J^^^^%^ 




f. 


^ 


^ 


<^\7 






-^HS 


K5\ 






^^ 






^^4/" 1" 










--^7^<^^^^^ 




4Q 




a: 


\<^ ^Vx 



Fig. 95. 




42 



DESCRIPTIVE GEOMETRY 



57. To determine the point in which a line 
pierces a plane. 

Method. 1. Pass an auxiliary plane 
through the line to intersect the given plane. 
2. Determine the line of intersection between 
the given and auxiliary planes. 3. The re- 
quired point will lie at the intersection of the 
given line and the line of intersection between 
the given and auxiliary planes. 

There may be four cases as follows: 

Case 1. When any auxiliary plane contain- 
ing the line is used. 

Case 2. When the horizontal or vertical 
projecting plane of the line is used. 

Case 3. When the given line is parallel to 
P, thus necessitating the use of an auxiliary 
profile plane containing the line. 

Case 4. When the given plane is defined by 
two lines which are not the traces of the plane. 

58. Case 1. Whenany auxiliary plane con- 
taining the line is used. 

Construction. Fig. 97. Let A be the 
given line and JV the given plane. Through 
line A pass any auxiliary plane Z (Art. 29, page 



20) intersecting plane iV in line C (Art. 50, 
page 35). Since lines A and (7 lie in plane Z^ d 
is their point of intersection, and since (7 is a 
line of plane JV^ point d is common to both line 
A and plane iV; hence, their intersection. 

59. Case 2. When the horizontal or verti- 
cal projecting plane of the line is used. 

Construction. Fig. 98 represents the 
same line A and plane iVof the previous figure. 
Pass the horizontal projecting plane X of line 
A (Art. 8, page 7), intersecting plane iV in 
line (Art. 50, page 35). Lines A and C in- 
tersect in point c?, the required point of pierc- 
ing of line A and plane A^. 

Fig. 99 is the solution of the same problem 
by the use of plane Y", the vertical projecting 
plane of line A. 

60. Case 3. When the given line is parallel 
to P, thus necessitating the use of an auxiliary 
profile plane containing the line. 

Construction. Fig. 100. Let A" be the 
given plane and line ab, parallel to P, the given 
line. Pass an auxiliary profile plane P through 
the given line ab^ intersecting plane A" in the 



PIERCING OF THE LINE AND PLANE 



43 



line C, having C^ for its profile projection profile projection of the required point of 

(Art. 54, page 38). Also determine a^h^, the piercing of line ah and plane N. The vertical 

profile projection of the given line (Art.- 21, and horizontal projections of this point are 

page 14). C^ and a^h^ intersect in dp, the d"" and d^. 




Fip-. 97. 




Fig. 98. 





<- 


XI 


\ 


/^\ \ 


a; 


Fig. 99.^ 


\ 





h 
a 


-/-—ya^ 




^ 5: 


V-*/ 


^ 


ct 


--V" 1 




i 


1^1 \ 1 

1 1 \ 1 


"^\ 


b° 


/ / 




^""-^ 


y ' 




^ 






a" 


^^^ 



Fig. 100. 



44 



DESCRIPTIVE GEOMETRY 



6i. Case 4. When the given plane is de- 
fined by two lines which are not its traces. 

Construction. Figs. 101 and 102. Let 
the given plane be defined by lines B and i>, 
and let A be the given line intersecting this 
plane at some point to be determined. Tliis 
case should be solved without the use of the 
traces of any plane. Pass the horizontal pro- 
jecting plane of line A intersecting line B in 
point g, and line D in point/, and consequently 
intersecting the plane of lines B and D in line 
ef. Because this horizontal projecting plane 
of line A is perpendicular to H all lines lying 
in it will have their horizontal projections 
coinciding, and ^y* will be the horizontal pro- 
jection of the line of intersection between the 
auxiliary plane and the plane of lines B and 
D. Draw the vertical projection of this line 
through the vertical projections of points e and 
/; its intersection with A^\ at d"^ will be the 
vertical projection of the required point of 
piercing of line A with the plane of lines B 
and B. 

The same result may be obtained by the use 



of the vertical projecting plane of line A. 

62. If a right line is perpendicular to a plane, 
the projections of that line will be perpendicular 
to the traces of the plane. In Figs. 103 and 
104 HN and VN are the traces of a plane to 
which line A is perpendicular. The horizontal 
projecting plane of line A is perpendicular to 
H by construction; it is also perpendicular to 
plane iV because it contains a line, A^ perpen- 
dicular to N\ therefore, being perpendicular 
to two planes it is perpendicular to their line 
of intersection, HN. But ffJVis perpendicular 
to every line in the horizontal projecting plane 
of line A which intersects it, and, therefore, 
perpendicular to A'\ Q.E.D. 

In like manner VN may be proved to be 
perpendicular to A . The converse of this 
proposition is true. 

63. To project a point on to an oblique plane. 
Principle. The projection of a point on 

any plane is the intersection with that plane 
of a perpendicular let fall from the point to 
the plane. 

Method. 1. From the given point draw 



A PERPEXDICUL.IR TO A PLANE 



45 




Fig. 101 




a perpendicular to tlie given plane (Art. 62, 
page 44). 2. Determine the point of piercing 
of this perpendicular and the plane (Art. 57, 
page 42). This point of piercing is the re- 
Cjuired projection of the given point upon the 
giA'en oblique plane. 

64. To project a given line on to a given 
oblique plane. 

Method. If the line is a right line, project 
two of its points (Art. 63, page 44). If the 
line is curved, project a sufficient number of 
its points to describe the curve. 

65. To determine the shortest distance from 
a point to a plane. 

Pkln'CIPLE. The shortest distance from a 
point to a plane is the perpendicular distance 
from the point to the plane. 

Method. 1. From the given point draAv a 
perpendicular to the given plane (Art. 62. 
page 44). 2. Determine the point of piercing 
of the perpendicular and the plane (Art. 57, 
page 42). 3. Determine the true length of 
the perpendicular between the point and the 
plane (Art. 39. page 27). 



46 



DESCRIPTIVE GEOMETRY 



66. Shades and Shadows. The graphic 
representation of objects, especially those of 
an architectural character, may be made more 
effective and more easily understood by draw- 
ing the shadow cast by the object. 

When a body is subjected to rays of light, 
that portion which is turned away from the 
source of light, and which, therefore, does not 
receive any of its rays, is said to be in shade. 
See Fig. 105. When a surface is in light and 
an object is placed between it and the source 
of light, intercepting thereby some of the rays, 
that portion of the surface from which light is 
thus excluded is said to be in shadow. That 
portion of space from which light is excluded 
is called the umbra or invisible shadow. 

(a) The umbra of a point in space is evi- 
dently a line. 

(5) The umbra of a line is in general a 
plane. 

((?) The umbra of a plane is in general a 
solid. 

(d^ It is also evident from Fig. 105 that 
the shadow of an object upon another object 



is the intersection of the umbra of the first 
object with the surface of the second object. 

The line of separation between the portion of 
an object in light and the portion in shade is 
called the shade line. It is evident from Fig. 
105 that the shade line is the boundary of the 
shade. It is also evident that the shadow of 
the object is the space inclosed by the shadow 
of its shade line. 

The source of light is supposed to be at an 
infinite distance; therefore, the rays of light 
will be parallel and will be represented by 
straight lines. The assumed direction of the 
conventional ray of light is that of the diag- 
onal of a cube, sloping downward, backward, 
and to the right, the cube being placed so that 
its faces are either parallel or perpendicular to 
F, H, and P (Fig. 106). This ray of light 
makes an angle of 35° 15' 52'^ with the coordi- 
nate planes of projection, but from Figs. 106 
and 107 it will be observed that the projections 
of the ray are diagonals of squares, and hence, 
they make angles of 45° with (xL. 

Since an object is represented by its projec- 



SHADES AND SHADOWS 



47 



tions, the ray of light must be represented by 
its projections. 

An object must be situated in the first 
quadrant to cast a shadow upon both l^and H. 

67. To determine the shadow of a point on 
a given surface pass the umbra of the point, or 
as is generally termed, pass a ray of light 
through the point and determine its intersec- 
tion with the given surface by Art. 24, page 
16, or by Art. 57, page 42, according as the 
shadow is required on a coordinate or on an 
oblique plane. 

68. To determine the shadow of a line upon 
a given surface it is necessary to determine 
the intersection of its umbra with that surface. 
If the line be a right line, this is generally best 
accomplished by finding the shadows of each 
end of the line and joining them. If the line 
be curved, then the shadows of several points 
of the line must be obtained. 

In Figs. 108 and 109 cb is an oblique line 
having one extremity, c, in H. By passing 
the ray of light through h and locating its 
horizontal trace, the shadow of h is found to 



fall upon H at h'^. Since c lies in H, it is its 
own shadow upon H\ therefore, the shadow of 
line ch upon H is c'^h^^. 




48 



DESCRIPTIVE GEOMETRY 



69. To determine the shadow of a solid upon 
a given surface it is necessary to determine the 
shadows of its shade lines. When it is diffi- 
cult to recognize which lines of the object are 
its shade lines, it is well to cast the shadow of 
every line of the object. The outline of these 
shadows will be the required result. 

Fig. 110 I represents an hexagonal prism 
located in the first quadrant with its axis per- 
pendicular to H. Its shadow is represented 
as falling wholly upon FJ as it would appear 
if 5^ were removed. Fig. Ill represents the 
prism when the shadow falls wholly upon H^ 
as it would appear if Fwere removed. Fig. 
112 represents the same prism when both V 
and H are in position, a portion of the shadow 
falling upon F", and a portion falling upon H. 
It is now readily observed that the shade lines 
are he, cd, de, ep, ps, sg, gk, and M, and that 
the shadow of the prism is the polygon in- 
closed by the shadows of these shade lines. 

When an object is so located that its 
shadow falls partly upon V and partly upon 
H, it is g-enerally best to determine first, its 



complete shadow upon both V and H, and to 
retain only such portions of the shadow as 
fall upon ]^ above H, and upon ^before V. 

Fig. 113 represents a right hexagonal pyra- 
mid resting upon an oblique plane N and 
having its axis perpendicular to that plane. 
Its shadow has been cast upon N, and the 
construction necessary to determine the 
shadow of one point, a, has been shown (Art. 
59, page 42). 

From the foregoing figures these facts will 
be observed : 

If a point lies on a plane, it is its own shadow 
upon that plane. See point C^ Figs. 108, 109, 
and the apex of the pyramid. Fig. 113. 

If a line is parallel to a plane, its shadow 
upon that plane will be parallel, and equal in 
length, to the line in space. Lines hk and ep, 
Fig. 110, are parallel to V\ lines he and sp, ed 
and gs, de and gk^ Fig. Ill, are parallel to H, 
and the sides of the hexagon, Fig. 113, are 
parallel to N. 

If two lines are parallel, their shadows are 
parallel. Observe that the shadows of the 



SHADES AND SHADOWS 



49 



opposite sides of the hexagons, and of the lat- 
eral edges of the prisms, are parallel. 

If a line is perpendicular to a coordinate 
plane, its shadow on that plane will fall on 
the projections of the rays of light passing 
through it. Lines hk and ep. Fig. Ill, fulfill 
this condition. 



fl"6V" c"e"</" 




a" b"f" Ce" d 



Fig. 113. 
a" b" r Ce'^d" 




H Fig. 112. 



50 



DESCRIPTIVE GEOMETRY 



70. Through a point or line to pass a plane 
having a defined relation to a given line or 
plane. There may be five cases, as follows : 

Case 1. To pass a plane through a given 
point parallel to a given plane. 

Case 2. To pass a plane through a given 
point perpendicular to a given line. 

Case 3. To pass a plane through a given 
point parallel to two given lines. 

Case 4. To pass a plane through a given 
line parallel to another given line. 

Case 5. To pass a plane through a given 
line perpendicular to a given plane. 

In cases 1 and 2 the directions of the re- 
quired traces are known. In cases 3, 4, and 
5 two lines of the required plane are known. 

71. Case I. To pass a plane through a given 
point parallel to a given plane. 

Pkikciple. Since the required plane is to 
be parallel to the given plane, their traces will 
be parallel, and aline through the given point 
parallel to either trace will determine the 
plane. 

Method. 1. Through the given point 



pass a line parallel to one of the coordinate 
planes and lying in the required plane (Art. 
9, page 8). 2. Determine the trace of this 
line, thus determining one point in the re- 
quired trace of the plane. 3. Draw the 
traces parallel to those of the given plane. 

Construction. Fig. 114. Let JV be the 
given plane and b the given point. Through 
b pass line A parallel to V and in such a di- 
rection that it will lie in the required plane, 
A^ being, parallel to GrL and A^ parallel to VN 
(Art. 35, page 22). Determine d, the hori- 
zontal trace of line A, and through d^ draw 
HS^ the horizontal trace of the required plane, 
parallel to ^iV. VjS will be parallel to A\ 

72. Case 2. To pass a plane through a given 
point perpendicular to a given line. 

Principle. Since the required plane is to 
be perpendicular to the given line, the traces of 
the plane will be perpendicular to the projec- 
tions of the line (Art. 62, page 44). Hence, 
a line drawn through the given point parallel 
to either coordinate plane, and lying in the 
required plane, will determine this plane. 



RELATION BETWEEN LINES AND PLANES 



51 




Fig. I 16. 



Method. 1. Through the given point pass 
a line parallel to one of the coordinate planes 
and lying in the required plane. 2. Deter- 
mine the trace of this line, thus determining 
one point in the required trace of the plane. 
3. Draw the traces j)erpendicular to the pro- 
jections of the given line. 

Construction. Fig. 115. Through point 
h draw line O parallel to Fand lying in the 
required plane S. C^ will be perpendicular 
to A'' (Art. 62, page 44). HS and VS are 
the required traces. 

73. Case 3. To pass a plane through a given 
point parallel to two given lines. 

Principle. The required plane will con- 
tain lines drawn through the given point 
parallel to the given lines. 

Method. 1. Through the given point pass 
two lines parallel to the two giv^en lines. 
2. Determine the plane of these lines (Art. 
30, page 20). 

Construction. Fig. 116. Given lines A 
and B and point h. Through point h pass lines 
C and I) parallel respectively to lines A and B. 
S, the plane of these lines, is the required plane. 



52 



DESCRIPTIVE GEOMETRY 



74. Case 4. To pass a plane through a given 
line parallel to another given line. 

Principle. The required plane will con- 
tain one of the given lines and a line inter- 
secting it and parallel to the second. 

Method. 1. Through any point of the 
given line pass a line parallel to the second 
given line. 2. Determine the plane of these 
intersecting lines (Art. 30, page 20). 

75. Cases. To pass a plane through a given 
line perpendicular to a given plane. 

Principle. The required plane will con- 
tain the given line and a line intersecting this 
line and perpendicular to the given plane. 

Method. 1. Through any point of the 
given line pass a line perpendicular to the 
given plane. 2. Determine the plane of these 
lines. 

Construction. Fig. 117. Given line A 
and plane iV. Through any point of Hue A 
pass line perpendicular to plane JV (Art. 
62, page 44). Determine S^ the plane of lines 
A and O (Art. 30, page 20). 

*76. Special conditions and methods of Art. 70. 

Case 1.* To pass a plane through a given 



point parallel to a given plane. 

Fig. 118 illustrates a condition in which the 
given plane iVis parallel to GL, which neces- 
sitates the use of an auxiliary profile plane. 
Through the given point b pass the profile 
plane P. Determine b^ and PA^, and through 
b^ draw PjS^ the profile trace of the required 
plane, parallel to PiV, whence Fas' and HjS^ the 
traces of the required plane, are determined. 

This condition may also be solved, Fig. 119, 
by passing a line A through the given point 
b, parallel to ani/ line C of the given plane 
JST. Through the traces of line A the traces 
of the required plane S are drawn parallel to 
ffiVand VN^ respectively. 

This method is applicable to all conditions 
of Case 1. 

77. Case 2.* To pass a plane through a 
given point perpendicular to a given line. 

Fig. 120 illustrates a condition in which the 
projections of the given line are perpendicular 
to GrL ; hence, the traces of the required plane 
will be parallel to GrL. Let ac be the given 

* The numbers of the cases are the same as those in 
Art. 70, page 50. 



RELATION BETWEEN LINES AND PLANES 



53 






vs 


/I 


VN 


y^ ! 


^i 


1 1 / 
1 0/ 


\/ HN 


y 


HS 



Fig. I 19. 




Fig. 121 




Fig. 122 



line and h the given point. Pass an auxiliary 
profile plane P, and determine a^c^, the pro- 
file projection of the given line, and 5^, the 
profile projection of the given point. Through 
h^ draw Pas', the profile trace of the required 
plane, perpendicular to o^cp, whence VS and 
HS are determined. 

78. Case 5.* To pass a plane through a 
given line perpendicular to a given plane. 

Fig. 121 illustrates a condition in which the 
given line ah is parallel to P. This may be 
solved by the use of an auxiliary profile plane, 
or by the following method: Through two 
points of the given line ah pass auxiliary lines 
O'and I) perpendicular to the given plane iV, 
and determine S, the plane of these parallel 
lines. Then S is the required plane contain- 
ing line ah and perpendicular to plan^ N. 

Fig. 122 illustrates a condition in which the 
given plane N is parallel to GL. This solu- 
tion is identical with that of Art. 75, page 52, 
save that to determine the traces of the auxil- 
iary line B, an auxiliary profile plane P has 
been used. Since B is to be perpendicular to 
iV, B^ must be perpendicular to PN. 



54 



DESCRIPTIVE GEOMETRY 



79. To determine the projections and true 
length of the line measuring the shortest distance 
between two right lines not in the same plane. 

Principle. The shortest distance between 
two right lines not in the same plane is the 
perpendicular distance between them, and 
only one perpendicular can be drawn termi- 
nating in these two lines. 

Method. 1. Through one of the given 
lines pass a plane parallel to the second given 
line. 2. Project the second line on to 
the plane passed through the first. 3. At the 
point of intersection of the first line and the 
projection of the second erect a perpendicular 
to the plane. This perpendicular will inter- 
sect the second line. 4. Determine the true 
length of the perpendicular, thus obtaining the 
required result. 

Construction. Figs. 123 and 124. Given 
lines A and B. Pass the plane S through 
line A parallel to line B (Art. 74, page 52). 
Project line B on to plane S at B^, using the 
auxiliary line D perpendicular to plane S and 
intersecting it at point k (Art. 64, page 45). 



Since line B is parallel to plane S^ B^ will be 
parallel to B ; hence, their projections are 
parallel (Art. 13, page 8). B^ intersects A 
at n, at which point erect the required line JE' 
perpendicular to plane aS* (Art. 62, page 44), 
intersecting line B at 0. Determine the true 
length of line U (not shown in the figure) by 
Art. 39, page 27. 

80. To determine the angle between a line 
and a plane. 

Principle. The angle which a line makes 
with a plane is the angle which the line makes 
with its projection cm the plane, or the com- 
plement of the angle which the line makes 
with a perpendicular which may project any 
point of the line on to the plane. The line in 
space, its projection on the plane, and the 
projector, form a right-angled triangle. 

Method. 1. Through any point of the 
given line drop a perpendicular to the given 
plane. 2. Determine the angle between this 
perpendicular and the given line. The angle 
thus determined is the complement of the 
required angle. 



ANGLE BETWEEN LINE AND PLANE 



55 



Construction. Fig. 125. Given line A 
and plane W. Through any point o, of line .4, 
pass line B perpendicular to plane iV (Art. 
62, page 44). Determine one of the traces of 
the plane of lines A and B, as VS. Revolve 
lines A and B into J^ about VS as an axis, as 
at A' and B' (Art. 43, page 31). Then 
angle d^c'e'" is the true size of the angle be- 
tween lines A and B, and its complement, 
e'"c% is the required angle between line A and 
plane iV(Art. 44, page 31). 



B 



Fig. 126 illustrates a condition in which the 
given plane iVis parallel with GL ; hence, the 
auxiliary line B^ perpendicular to iV, is parallel 
to P, thus requiring a profile projection to 
determine its traces. VS is the vertical trace 
of the plane of lines A and B (its other trace 
is not necessary), and angle e"c'f is the required 
angle between line A and plane iV. 

8i. To determine the angle between a line 
and the coordinate planes. 

Principle. This is a special case of Art. 
80, page 54. 




Fig. 124 



56 



DESCRIPTIVE GEOMETRY 



1st Method. 1. Revolve the line about its 
liorizontal projection as an axis to obtain the 
angle which it makes with H, 2. Revolve 
the line about its vertical projection as an 
axis to obtain the angle which the line makes 
with V. 

CoKSTKUCTiON. Fig. 127 represents the 
given line A when revolved into K and 
measures a, the true angle which the line 
makes with H. Fig. 127 also represents the 
line as revolved into Fand measuring yS, the 
true angle which the line makes with V 
(Art. 41, page 28). 

2kd Method. 1. Revolve the line parallel 
to Fto obtain the angle which the line makes 
with 1?. 2. Revolve the line parallel to ^ to 
obtain the angle which the line makes with V. 

Construction. Fig. 128 represents the 
given line A when revolved parallel to Fand 
measures ot, the true angle which the line 
makes with H. Fig. 128 also represents the 
line A as revolved parallel to IT and measur- 
ing /3, the true angle which the line makes 
with V. ' 



82. To determine the projections of a line of 
definite length passing through a given point 
and making given angles with the coordinate 
planes. 

Principle. ^ This is the converse of Art. 
81, page 55^ and there may be eiglit solutions; 
but the sum of the angles, a and /3, which the 
line makes with the coordinate planes, cannot 
be greater than 90°. 

Construction. Fig. 128. Let it be re- 
quired to draw a line through point h having 
a length equal to x and making angles of a 
and /3 with II and F, respectively. Through 
h^ draw h^cl equal to x and making angle a 
with GL; also b''c'l parallel to GL. These 
projections will represent the revolved posi- 
tion of the line, making its required angle 
a with IT. Similarly draw the projections 
h^c'2, b^'cl to show the revolved position and 
required angle yS with F. 

In counter-revolution about a vertical axis 
through 5, all possible horizontal projections 
of line A will be drawn from b^ to a line 
through (?2 and parallel to GfL. But all pos- 



ANGLE BETWEEN PLANES 



57 



sible horizontal projections must be drawn 
from h^ to the arc e\e''\ hence, tf^ the intersec- 
tion of this arc and the parallel through 4, 
will be the horizontal projection of the other 
extremity of the required line. Since the 
vertical projection of this point must also lie 
in the parallel to GL through c^^, the line is 
definitely determined. 




83. To determine the angle between two 
planes. 

Principle. If a plane be passed perpendic- 
ular to the edge of a diedral angle, it intersects 
the planes of the diedral in lines, the angle 
between which is known as the plane angle of 
the diedral. The diedral angle between two 
planes is measured by its plane angle. 

Method. 1. Determine the plane angle 
of the given diedral angle by passing an 
auxiliary plane perpendicular to the line of 
intersection between the two given planes, 
and, therefore, perpendicular to each of these 
given planes. 2. Find the lines of intersec- 
tion between this auxiliary plane and the 
given planes. 3. Determine the true size of 
this plane angle. 

84. Case 1. When it is required to de- 
termine the angle between any two oblique 
planes. 

Construction. Figs. 129 and 130. The 
lettering of these two figures is identical 
although the diedral angles differ. 



58 



DESCRIPTIVE GEOMETRY 



Pass the auxiliaiy plane S perpendicular to 
A, the line of intersection between the planes 
iV and L. Only one trace of plane S is neces- 
sary for the solution of the problem and in 
this example the horizontal trace has been 
selected. Then ^aS' must be perpendicular 
to A^ (Art. 62, page 44). Its intersection 
with the horizontal traces of the given planes 
will determine points / and e, one in each 
line of intersection between the auxiliary and 
given planes. The point d, which is com- 
mon to both lines of intersection, may be 
obtained as follows: The horizontal project- 
ing plane of A cuts the auxiliary plane in 
do, a line perpendicular to A. By revolving 
A into IT the revolved position of dc may be 
drawn, as at d^e, and by counter-revolution 
point d obtained. Next revolve de and df 
into H to measure the angle between them, 
which is the required diedral angle ed''f. 

One example of this problem which is com- 
monly met in practice is here shown in solu- 
tion in Fig. 131. Let N and L be the given 
planes, A their line of intersection, and plane 



S the plane passed perpendicular to A. Then 
angle e^d"f^ is the true size of the angle be- 
tween planes N and L. 

85. Another method for determining the 
size of the angle between two planes is illus- 
trated by Fig. 132. This is a pictorial repre- 
sentation of two intersecting planes, iV^and R, 
From any point in space, as point a, two lines 
are dropped perpendicular, one to each plane. 
The angle between these perpendiculars is the 
measurement of the angle between planes N 
and i^, or its supplement. 

86. Case 2. When it is required to deter- 
mine the angle between an inclined plane and 
either coordinate plane. 

Construction. Figs. 133 and 134. Let 
it be required to determine the true size of 
the angle between planes iVand H. Pass the 
auxiliary plane JT perpendicular to both iVand 
H^ and, therefore, perpendicular to their line of 
intersection, HN. Then HX and VX are 
perpendicular respectively to ffiV^and GrL. 

The angle between the lines ah and HX, in 
which plane X intersects planes N and H re- 



ANGLE BETWEEN PLANES 



59 



spectively, is the required angle, the true size mum inclination of plane iVwith H. 

of which, «, is determined by revolving the Suppose it is required to determine the true 

triangle containing it into F about FX as an size of the angle between planes iV and F. 

axis, or into ^ about HX as an axis. Determine the line of maximum inclination of 

The line ah is known as the line of maxi- plane N with F, and its angle with F. 



^^ 




60 



DESCRIPTIVE GEOMETRY 



87. To determine the bevels for the correct 
cuts, the lengths of hip and jack rafters, and 
the bevels for the purlins for a hip roof. 

Fig. 135 represents an elevation and plan 
of a common type of hip roof having a pitch 

equal to -— - and a width of c"h'". ah is the 

center line of the hip rafter ; ^ is a cross sec- 
tion of the ridge ; F^ K, Z, M, are jack rafters. 

Hip. To find the true length of the hip 
rafter eh^ and the following angles: 

Down cut, 1: The intersection of the hip 
with the ridge. 

Heel cut, 2: The intersection of the hip 
with the plate. 

Side cut, 3: Intersection of the hip with 
the ridge. 

Top bevel of hip, 4. 

The true length of the hip eh may be ob- 
tained by revolving it parallel to V, as at e-J?'"^ 
or parallel to H^ as at e^h'^. The down cut 
bevel, 5, is obtained at the same time. The 
bevel of the top edge of hip is found by pass- 
ing a plane perpendicular to ah intersecting 



the planes of the side and end roofs. HZ is 
the horizontal trace of this plane, and the 
bevel, 4, is obtained as in Art. 84, page 57. 

Jack Rafters. The down cut bevel, 5, 
and the heel cut bevel, ^, of the jack rafters 
are shown in their true values in the elevation ; 
and the side cut, 7, is shown in the plan. 
The true lengths of the jack rafters are ob- 
tained by extending the planes of their edges 
to intersect the revolved position of the hip 
rafter, as at n^o^. 

Purlin. It is required to determine the 
down cut, 8, side cut, 9, and angle between side 
and end face of purlin, 10. To obtain the 
down cut revolve side face parallel to ^and 
the true angle, 8, will be obtained. Similarly, 
the side cut made on the top or bottom face 
is obtained by revolving that face parallel to 
H^ the true angle being indicated by bevel, 9. 
In order to obtain the angle between the side 
and end faces, the planes of which are indi- 
cated in the figure by S and R, find the inter- 
section between these planes and determine 
the angle as in Art. 84, page 57. 



ANGLE BETWEEN PLANES 



61 




Fig. 135. 



62 



DESCRIPTIVE GEOMETRY 



88. Given one trace of a plane, and the angle 
between the plane and the coordinate plane, to 
determine the other trace. 

There must be two cases, as follows : 

Case 1. Given the trace on one coordinate 
plane and the angle which the plane makes 
with the same coordinate plane. 

Construction. Fig. 136. Let ITT be the 
given trace and a the given angle between T 
and H. Draw A^ perpendicular to IIT, it being 
the horizontal projection of the line of maxi- 
mum inclination with H (Art. 86, page 58). 
If this line be revolved into IT^ it will make 
the angle a with J.^ and d' will be the resolved 
position of the vertical trace of the line of 
maximum inclination with If. The vertical 
projection of this trace must lie on the vertical 
trace of the horizontal projecting plane of A 
and at a distance from GrL equal to d'^d'. 
Therefore, d^ will be a point in VT^ the re- 
quired trace. 

There may be two solutions, as VT may be 
above or below GrL. 

89. Case 2, Given the trace on one coor- 



dinate plane and the angle which the plane 
makes with the other coordinate plane. 

Construction. Fig. 137. Let RT he the 
given trace and jS the given angle between T 
and V, 

Consider B^ the line of maximum inclina- 
tion with FJ as revolved about the horizontal 
trace of its vertical projecting plane and mak- 
ing an angle, yS, with GL. d' will be the 
revolved position of the vertical trace of B. 
From g, with radius ed\ describe arc d'd'". VT, 
the required trace, will be tangent to this arc. 
There may be two solutions, as VT may be 
above or below GL. 

90. To determine the traces of a plane, 
knowing the angles which the plane makes with 
both coordinate planes. 

Construction. Fig. 138. Let it be re- 
quired to construct the traces of plane T, mak- 
ing an angle ^ with l^and angle a with IT. 
The sum of a and /S must not be less than 
90° nor more than 180°. Conceive the re- 
quired plane T as being tangent to a sphere, 
the center of which, 0, lies in CrL. From 



TO DETERMINE TRACES OF FLAXES 



63 




the point of tangency of the sphere and pldne 
T, conceive to be drawn the lines of maximum 
inclination with ^ and V. On revolving the 
line of maximum inclination with V, into E, 
about the horizontal axis of the sphere, as an 
axis, it will continue tangent to the sphere, 
as at A!, making an angle with GL equal to 
the required angle (between V and r), and 
its horizontal trace will lie in the axis of 
revolution at «\ its vertical trace lying in the 
circle described from e as a center, with a 
radius cm^ , Similarly revolve the line of 
maximum inclination with ^into V, as at B\ 
making an angle with GL equal to the re- 
quired angle (between ^and T), and its ver- 
tical trace will lie in the axis of revolution at 
^% its horizontal trace lying in the circle de- 
scribed from <? as a center, with a radius ce\ 
Then HT will contain aP- and be tangent to 
arc al'e^ and VT will contain h"^ and be tan- 
gent to arc 'k'm^. The traces must intersect 
GL in the same point. Both HT and VT 
may be either above or below GL. 

91. In Art. 48, page 34, reference was made 



64 



DESCRIPTIVE GEOMETRY 



to a fourth construction for counter-revolution. 
This construction involves the angle of maxi- 
mum inclination between the oblique and 
coordinate planes. Let it be required to draw 
the projections of a regular pentagon lying in 
plane N, Fig. 139, when its revolved position 
is known. Pass plane X perpendicular to 
both N and F, intersecting plane N in the 
line of maximum inclination with F, shown in 
revolved position as gf^. Since this line shows 
in its true length, all distances on plane N 
perpendicular to VN may be laid off on it. 
Then ge^' represents the distance from point e 
to VN, and e" will lie on a line through e" 
parallel to FiV. Likewise other points of the 
pentagon are determined. Also e"k is the 
distance from point e in space to F; hence, e^' 
will lie at a distance from CrL equal to e"k. 





^ 




•-- 


\ 
\ 


jy^ . 


^ 






\ 
\ 
\ 




5: 














\ 










\ 






f 




1 

1 


K rl" ! 


/ 


\ 




1 



C^> 



39. 



CtlAPTER III 

GENERATION AND CLASSIFICATION OF 
SURFACES 



92. Every surface may be regarded as hav- 
ing been generated by the motion of a line, 
which was governed by some definite law. 
The moving line is called the generatrix, and 
its different positions are called elements of 
the surface. Any two successive positions of 
the generatrix, having no assignable distance 
between them, are called consecutive elements. 
The line which may direct or govern the gen- 
eratrix is called the directrix. 

93. Surfaces are classified according to the 
form of the generatrices, viz. : 

Ruled Surfaces, or such as may be gen- 
erated by a rectilinear generatrix. 



Double-curved Surfaces, or such as must 
be generated by a curvilinear generatrix. 
These have no rectilinear elements. 

The ruled surfaces are reclassified as devel- 
opable^ and nondevelopahle or warped surfaces. 

Plane 

Single-curved (developable). 
Cylinder 
Cone 
Surfaces <! Convolute 

Warped (nondevelopahle) 
^Douhle-curved (nondevelopable) 

94. Ruled Surfaces. A right line may move 
so that all of its positions will lie in the same 



' Ruled < 



65 



66 



DESCRIPTIVE GEOMETRY 



plane ; it may move so that any two consecu- 
tive elements will lie in the same plane ; or it 
may move so that any two consecutive ele- 
ments will not lie in the same plane. Thus, 
ruled surfaces are subdivided into three 
classes, as follows : 

Plane Surfaces : All the rectilinear ele- 
ments lie in the same plane. 

Single-curved Surfaces : Any two con- 
secutive rectilinear elements lie in the same 
plane, i.e. they intersect or are parallel. 

Warped Surfaces : No two consecutive 
rectilinear elements lie in the same plane, i.e. 
they are neither intersecting nor parallel. 

95. Plane Surfaces are all alike. The rec- 
tilinear generatrix may move so as to touch 
one rectilinear directrix, remaining always 
parallel to its first position ; so as to touch 
two rectilinear directrices which are parallel to 
each other, or which intersect; or it may re- 
volve about another riglit line to which it is 
perpendicular. 

96. Single-curved Surfaces may be divided 
into three classes, as follows : 



Cones: In which all the rectilinear ele- 
ments intersect in a point, called the apex. 

Cylinders : In which all the rectilinear 
elements are parallel to each other. The 
cylinder may be regarded as being a cone with 
its apex infinitely removed. 

Convolutes : In which the successive rec- 
tilinear elements intersect two and two, no 
three having one common point. 

97. Conical Surfaces are generated by the 
rectilinear generatrix moving so as always to 
pass through a fixed point, called the apex, 
and also to touch a given curve, called the 
directrix. Since the generatrix is indefinite in 
length, the surface is divided at the apex into 
two parts, called nappes. The portion of a 
conical surface usually considered is included 
between the apex and a plane which cuts all 
the elements. This plane is called the base of 
the cone and the form of its curve of inter- 
section with the conical surface gives a dis- 
tinguishing name to the cone, as, circular, 
elliptical, parabolic, etc. If the base of the 
cone has a center, the right line passing 



SINGLE CURVED SURFACES 



67 



through this center and the apex is called the 
axis of the cone (Fig. 140). 

A Right Cone is one having its base per- 
pendicular to its axis. 





Fig. 140. 



Fig. 141 



98. Cylindrical Surfaces. The cylinder is 
that limiting form of the cone in which the 
apex is removed to infinity. It may be gen- 



erated by a rectilinear generatrix which moves 
so as always to touch a given curved directrix, 
having all of its positions parallel. A plane 
cutting all the elements of a cylindrical sur- 
face is called its hase^ and the form of its curve 
of intersection with the surface gives a dis- 
tinguishing name to the cylinder, as in the 
case of the cone. If the base has a center, 
the right line through this center parallel to 
the elements is called the axis (Fig. 141). 

A cylinder may also be generated by a cur- 
vilinear generatrix, all points of which move in 
the same direction and with the same velocity. 

A Right Cylinder is one having its base 
perpendicular to its axis. 

99. Convolute Surfaces may be generated 
by a rectilinear generatrix which moves so as 
always to be tangent to a line of double cur- 
vature.* Any two consecutive elements, but 
no three, will lie in the same plane. Since 
there is an infinite number of lines of double 
curvature, a great variety of convolutes may 

* A line of double curvature is one of which no four 
consecutive points lie in the same plane. 



68 



DESCRIPTIVE GEOMETRY 



exist. One such form which may readily be 
generated is the helical convolute (Fig. 142). 
It is the surface generated by the hypotenuse 
of a right triangle under the following condi- 
tions : Suppose a right triangle of paper, or 
some other thin, flexible material, to be wrapped 
about a right cylinder, one leg of the triangle 
coinciding with an element of the cylinder. 
If the triangle be unwrapped, its vertex will 
describe the involute of the base of the cylin- 
der, and the locus of the points of tangency 
of its hypotenuse and the cylinder will be a 
helix, the hypotenuse generating the helical 
convolute. The convolute may also be re- 
garded as being generated by a rectilinear 
generatrix moving always in contact with the 
involute and helix as directrices, and making 
a definite angle with the plane of the involute. 
100- A Warped Surface is generated by a 
rectilinear generatrix moving in such a way 
that its consecutive positions do not lie in the 
same plane. Evidently there may be as many 
warped surfaces as there are distinct laws re- 
stricting the motion of the generatrix. 



Any warped surface may be generated by a 
rectilinear generatrix moving so as to touch 
two linear directrices, and having its consecu- 
tive positions parallel either to a given plane, 
called a plane director, or to the consecutive 
elements of a conical surface, called a cone 
director. 

loi. The following types, illustrated by 
Figs. 142 to 148, indicate the characteristic 
features of warped surfaces : 

Hyperbolic Paraboloid, Fig. 143. Two 
rectilinear directrices and a plane director, or 
three rectilinear directrices. 

Conoid, Fig, 144. One rectilinear and one 
curvilinear directrix and a plane director. 

Cylindroid, Fig. 145. Two curvilinear 
directrices and a plane director. 

Right Helicoid, Fig. 146. Two curvi- 
linear directrices and a plane director. 

Oblique Helicoid, Fig. 147. Two curvi- 
linear directrices and a cone director. 

Hyperboloid of Revolution, Fig. 148. 
Two curvilinear directrices and a cone direc- 
tor, or three rectilinear directices. 



WARPED SURFACES 



69 




Fig. 147. 



70 



DESCRIPTIVE GEOMETRY 



102. A Surface of Revolution, Fig. 149, is 
tlie locus of any line, or generatrix, the posi- 
tion of which remains unaltered with reference 
to a fixed right line about which it revolves. 
This fixed right line is called the axis of revo- 
lution. A circle of the surface generated by 
any point of the generatrix is called 'd parallel., 
and planes perpendicular to the axis will cut 
the surface in parallels. Any plane contain- 
ing the axis of revolution is called a merid- 
ian plane^ and the line cut from the surface 
by this plane is called a meridian line. All 
meridian lines of the same surface are obvi- 
ously identical, and any one of them may be 
considered as a generatrix. That meridian 
plane which is parallel to a coordinate plane 
is called the principal meridian. 

If the generatrix be a right line lying in 
the same plane as the axis, it will either be 
parallel with it or intersect it ; in the former 
case the surface generated will be a cylinder, 
in the latter, a cone, and these are the only 
single-curved surfaces of revolution. 

If the generatrix does not lie in the same 



plane with the axis, the consecutive positions 
are neither parallel nor intersecting. The 
surface must then be warped and its meridian 
line will be an hyperbola. This is the only 
warped surface of revolution. It may also be 
generated by revolving an hyperbola about its 
conjugate axis, and is known as the hyperbo- 
loid of revolution of one nappe (Fig. 148). 

103. Double-curved Surfaces. With the ex- 
ception of the cylinder, cone, and hyperboloid 
of revolution all surfaces of revolution are of 
double curvature. They are infinite in num- 
ber and variety. Representative types are : 

TfiE Sphere: Generated by revolving a 
circle about its diameter. 

The Prolate Spheroid: Generated by 
revolving an ellipse about its major axis. 

The Oblate Spheroid : Generated by 
revolving an ellipse about its minor axis. 

The Paraboloid : Generated by revolv- 
ing a parabola about its axis (Fig. 149). 

The Hyperboloid of Two Nappes : 
Generated by revolving an hyperbola about 
its transverse axis. 



DOUBLE-CURVED SURFACES 



71 



The Torus (annular or not): Generated 
by revolving a circle about a line of its plane 
other than its diameter. Fig. 150 illustrates 
an annular torus. 



The Double-curved Surface of Traxs- 
POSiTiox — Serpentine : Generated by a 
sphere the center of which moves along an 
helix (Fig. 151). 






PARABOLOID 
Fig. 149. 



TORUS {ANNULAR). 

Fis:. 150. 




SERPENTINE 
Fig. 151. 



CHAPTER IV 



TANGENT PLANES 



104. A plane is tangent to a single-curved 
surface when it contains one, and only one 
element of that surface. The two lines com- 
monly determining a tangent plane are the tan- 
gent element and a tangent to the surface at 
some point in this element. If this second 
line lies in one of the coordinate planes it will 
be a trace of the tangent plane. 

In Fig. 152 point d is on the surface of a 
cone to which a tangent plane is to be drawn. 
The line drawn through point d and the apex 
of the cone will be the element at which the 



plane is to be tangent and, therefore, one line 
of the tangent plane. If a second line, ck^ be 
drawn through point d, and tangent to any 
section of the cone containing this point, it 
will be a second line of the tangent plane. 
The traces of these lines will determine ^aS' 
and VS^ the traces of the required tangent 
plane. If the base of the single-curved sur- 
face coincides with one of the coordinate 
planes, as in Fig. 152, hk can be used for 
the tangent line, thus determining the hori- 
zontal trace directly. 



72 



TANGENT PLANES 



73 




105. One projection of a point on a single- 
curved surface being given, it is required to 
pass a plane tangent to the surface at the ele- 
ment containing the given point. 

Principle. The tangent plane will be de- 
termined by two intersecting lines, one of 
which is the element of the surface on which 
the given point lies, and the second is a line 
intersecting this element and tangent to the 
single-curved surface, preferably in the plane 
of the base. 

Method. 1. Draw the projections of the 
element containing the given point. 2. In 
the plane of the base draw a second line 
tangent to the base at the tangent element. 
8. Determine the plane of these lines. 

If the plane of the base coincides with one 
of the coordinate planes, the tangent line will 
be one of the traces of the required tangent 
plane. 

Note. In this and the following problems 
the base of the single-curved surface is consid- 
ered as lying on one of the coordinate planes. 



74 



DESCRIPTIVE GEOMETRY 



Construction. Fig. 153. Let the single- 
curved surface be a cone which is defined by 
its projections, and the base of which lies in 
one of the coordinate planes; in this case in H. 
Let c" be the vertical projection of the given 
point. Through d" draw ^^ the vertical pro- 
jection of the tangent element. The vertical 
projection of the horizontal trace of this ele- 
ment is k\ which, being a point in the base, 
will be horizontally projected at k\ and k\^ thus 
making two possible tangent elements, E-^ and 
E^ ; also two possible locations for point c, 
and, therefore, two possible tangent planes. 
Since the base of the cone lies in H^ the 
horizontal traces, HS and HN of the tangent 
planes aS' and iV, will be tangent to the base 
at k\ ^^d ^1- The vertical traces of the 
tangent elements, V" and o", determine the 
necessary points in VS and F7V, the required 
vertical traces of the tangent planes. 

io6. To pass a plane tangent to a cone and 
through a given point outside its surface. 

Principle. Since all tangent planes con- 
tain the apex of the cone, the required plane 




PLAXE TAXGEXT TO COXE 



75 



must contain a line cIra^Yn through the apex 
of the cone and tlie given point. It must also 
have one of its traces tangent to the base of 
the cone, since this base is supposed to lie in 
a coordinate plane. 

Method. 1. Draw the projections of a 
line passing through the apex of the cone and 
the given point. 2. Determine its traces. 
3. Through that trace of the auxiliary line 
which lies in the plane of the base draw 
the trace of the required plane tangent to the 
base of the cone. 4. Draw the other trace of 
the plane through the other trace of the 
auxiliary line. There are two possible tan- 
gent planes. 

107. To pass a plane tangent to a cone and 
parallel to a given line. 

Pei:^sCIPLE. The tangent plane must con- 
tain the apex of the cone and a line through 
the apex parallel to the given line. 

^Method. 1. Through the apex of the 
cone draw a line parallel to the given line. 
2. Obtain the traces of this auxiliarv line. 



3. Draw the traces of the required tangent 
plane through the traces of the auxiliary line, 
making one of them tangent to the base of the 
cone. There are two possible tangent planes. 

108. To pass a plane tangent to a cylinder 
and through a given point outside its surface. 

PiuxciPLE. The tangent plane will be de- 
termined by two intersecting lines, one of 
Avliich is draAvn through the given point 
parallel to the elements of the cylinder, and 
the second is drawn tangent to the cylinder 
from some point in the first line, preferably in 
tlie plane of the base. 

Method. 1. Through the given point 
draw the projections of a line parallel to the 
elements of the cylinder. 2. Determine the 
traces of this auxiliary line. 3. Through 
that trace of the auxiliary line which lies in 
the plane of the base of the cylinder draw one 
trace of the required plane tangent to the base. 

4. Through the other trace of the auxiliary 
line draw the second trace of the tano-ent 
plane. There are two j^ossible tangent planes. 



76 



DESCRIPTIVE GEOMETRY 



Construction. Fig. 154. Let dJ" and a^ 
be the projections of a given point through 
which the plane is to be passed tangent to the 
cylinder. The base of the cylinder rests on H. 
B" and B^ are the projections of the auxiUary 
line drawn through point a parallel to the ele- 
ments of the cylinder. Through c?^ the hori- 
zontal trace of this line, HS and ffiV, the 
horizontal traces of the two possible tangent 
planes, may be drawn tangent to the base of 
the cylinder. The vertical traces of these 
planes, VS and FiV, must contain c", the verti- 
cal trace of the line B. The elements at which 
planes N and S are tangent are lines F and K^ 
respectively. 

109. To pass a plane tangent to a cylinder 
and parallel to a given line. 

Principle. The tangent plane must be 
parallel to a plane determined by the given 
line and a line intersecting it, which is parallel 
to the elements of the cylinder. 

Method. 1. Through any point of the 
given line draw a line parallel to the elements 




PLANE TANGENT TO CYLINDER 



77 




of the cylinder. 2. Determine the traces of 
the plane of these two lines. 3. Draw the 
traces of the tangent plane parallel to the 
traces of the auxiliary plane (Art. 18, page 
12), one of the traces being tangent to the 
base of the cylinder. There may be two tan- 
gent planes. 

Construction. Fig. 155. Through any 
point, ^, of the given line A, draw line B paral- 
lel to the elements of the cylinder. The tan- 
gent planes will be parallel to X, the plane of 
these lines. iV and S will be the required 
tangent planes. 

no. A plane is tangent to a double -curved 
surface when it contains one, and only one 
point of that surface. The two lines com- 
monly determining the tangent plane are the 
lines tangent to the meridian and parallel at 
the point of tangency (Art. 102, page 70). 

A Normal is the line perpendicular to the 
tangent plane at the point of tangency. 

A Normal Plane is any plane containing the 
normal. 



78 



DESCRIPTIVE GEOMETRY 



III. One projection of a point on the surface 
of a double -curved surface of revolution being 
given, it is required to pass a plane tangent to 
the surface at that point. 

Principle. Planes tangent to double- 
curved surfaces of revolution must contain 
the tangents to the meridian and parallel at 
the point of tangency. 

There may be two methods. 

1st Method. 1. Through the given point 
draw a meridian and a parallel (Art. 102, 
page 70). 2. Draw tangents to these curves 
at the given point. 3. Determine the plane 
of these tangents. 

Construction. Fig. 156. Given the ellip- 
soid with its axis perpendicular to H. A plane 
is required to be drawn tangent to the surface 
at the point having e^^ for its horizontal pro- 
jection. With f^e'^ as a radius, and /''' as a 
center, describe the circle which is the hori- 
zontal projection of the parallel through e. 
One of the possible vertical projections of the 
parallel is that portion of B^ lying within the 
ellipse. Project e'^ on to this line to obtain e". 



the vertical projection of the given point e. 
Through point e draw line B tangent to the 
parallel, and, therefore, in the plane of the 
parallel. That portion of A^ lying within 
the circle, which is the horizontal projection 
of the ellipsoid, will be the horizontal projec- 
tion of the meridian drawn through point e. 
Revolve this meridian line about fk as an axis 
until it coincides with the principal meridian 
(Art. 102, page 70), the vertical projection of 
which will be shown by the ellipse. The re- 
volved position of the vertical projection of 
point e will now be at e^^ and a line may be 
drawn tangent to the meridian at this point, 
its projections being A\, A^ Counter-revolve 
this meridian plane to determine the true posi- 
tion of the tangent line, shown by its projec- 
tions at A'\ A^. The traces of the tangent 
lines A and B will determine the traces of the 
required tangent plane. There are two possi- 
ble tangent planes. 

2nd Method. 1. Draw the projections of 
a cone tangent to the double-curved surface 
of revolution at the parallel passing through 



PLANE TANGENT TO ELLIPSOID 



79 




the given point. 2. Pass a plane tangent 
to the auxiliary cone at the element drawn 
through the given point (Art. 105, I3age 73). 

Construction. Fig. 156. A^ and J.^ are 
the projections of the element of a cone tan- 
gent to the ellipsoid and containing point e. 
A plane tangent to the cone at line A will be 
the required plane tangent to the ellipsoid 
at point e. If the vertical trace of line A 
lies beyond the limits of the paper, the direc- 
tion of FiV may be determined by observing 
that the trace of the required plane must be 
perpendicular to the normal D at the point e 
(Art. 110, page 77). There are two possible 
tangent planes. 

112. Through a point in space to pass a 
plane tangent to a double-curved surface of 
revolution at a given parallel. 

Method. 1. Draw a cone tangent to the 
double-curved surface of revolution at the 
given parallel. 2. Pass a plane through 
the given point tangent to the cone (Art. 
106, page 74). There are two possible tangent 
planes. 



80 



DESCRIPTIVE GEOMETRY 



113. To pass a plane tangent to a sphere at 
a given point on its surface. 

Method. This may be solved as in Art. 
Ill, or by the following method. 1. Through 
the given point draw a radius of the sphere. 
2. Pass a plane through the given point per- 
pendicular to the radius (Art. 72, page 50), 
and this will be the required plane. 

114. Through a given line to pass planes 
tangent to a sphere.* 

PRINCIPLE. Conceive a plane as passed 
through the center of the sphere and perpen- 
dicular to the given line. It will cut a great 
circle from the sphere and lines from the re- 
quired tangent planes. These lines will be 
tangent to the great circle of the sphere and 
intersect the given line at the point in which 
it intersects the auxiliary plane. The planes 
determined by these tangent lines and the 
given line will be the required tangent planes. 

* For a general solution of problems requiring the draw- 
ing of tangent planes to double-curved surfaces of revolu- 
tion, and through a given line, see Art. 160, page 122. 



Method. 1. Through the center of the 
sphere pass a plane perpendicular to the given 
line (Art. 72, page 50) and determine its 
traces. 2. Determine the point in which this 
plane is pierced by the given line (Art. 57, 
page 42). 3. Into one of the coordinate 
planes revolve the auxiliary plane containing 
the center of the sphere, the great circle cut 
from the sphere, and the point of intersection 
Avith the given line. 4. F'rom the latter 
point draw lines tangent to the revolved 
position of the great circle of the sphere. 
These lines will be lines of the required tan- 
gent planes. 5. Counter-revolve the auxil- 
iary plane containing the lines of the tangent 
planes. 6. Determine the planes defined by 
the given line and each of the tangent lines 
obtained by 4. These will be the required 
tangent planes. 

Construction. Fig. 157. The given line 
is A, and the center of the sphere is e. HX 
and VX are the traces of the auxiliary plane 
perpendicular to A and passing through e 



PLANE TANGENT TO SPHERE 



81 



(Art. 72, page 50). The point/ is that of the 
intersection of line A and plane X. Revolve 
plane X into the vertical coordinate plane and 
e' will be the revolved position of the center 
of the sphere, and f the revolved position of 
the point /. Draw the great circle of the 



sphere and the tangents C and B'. 

In counter-revolution these lines will be at 
C and B, intersecting A -dt f. C and A will 
be two intersecting lines of one of the required 
tangent planes, S, and the second plane, iV, will 
be determined by lines O and B. 




CHAPTER V 



INTERSECTION OF PLANES WITH SURFACES, 
AND THE DEVELOPMENT OF SURFACES 



of 



any 



115. To determine the intersection 
surface with any secant plane. 

General Method. 1. Pass a series of 
auxiliary cutting planes which will cut lines, 
straight or curved, from the surface, and right 
lines from the secant plane. 2. The inter- 
sections of these lines are points in the re- 
quired curve of intersection. 

This method is applicable alike to prisms, 
pyramids, cylinders, cones, or double-curved 



surfaces of revolution. The auxiliary cutting 
planes may be used in any position, but for 
convenience they should be chosen so as^ to 
cut the simplest curves from the surface, that 
is, straight lines or circles. 

With solids such as prisms, pyramids, 
single-curved, or other ruled surfaces, the 
above method consists in finding the intersec- 
tion of each element with the oblique plane by 
Art. 61, page 44. 



82 



INTERSECTION OF PLANE WITH PYRAMID 



83 



1 1 6. A tangent to the curve of intersection 

of a plane with a single-curved surface may be 
drawn by passing a plane tangent to the sur- 
face at the point assumed (Art. Ill, page 78). 
The line of intersection of the tangent plane 
Avith the secant plane will be the required 
tangent. 

117. The true size of the cut section may al- 
ways be found by revolving it into one of the 
coordinate planes, about a trace of the secant 
plane as an axis. 

118. A right section is the section cut from 
the surface by a plane perpendicular to the 
axis. 

119. The development of a surface is its 
true size and shape when spread open upon a 
plane. Only surfaces having two consecutive 
elements in the same plane can be developed, 
as only such surfaces can be made to coincide 
with a plane. Therefore, only single-curved 
surfaces, and solids bounded by planes, can be 
developed. Solids bounded by planes are 
developed by finding the true size and shape 



of each successive face. Single-curved sur- 
faces are developed by placing one element in 
contact with the plane and rolling the surface 
until every element has touched the plane. 
That portion of the plane covered by the sur- 
face in its revolution is the development of 
the surface. 

120. To determine the intersection of a plane 
with a pyramid. 

Peixciple. Since a pyramid is a solid 
bounded by planes, the problem resolves itself 
into determining the line of intersection be- 
tween two planes. Again, since the pyramid 
is represented by its edges, the problem still 
further resolves itself into determining the 
points of piercing of these edges with the 
plane. 

Method. 1. Determine the points in which 
the edges of the given pyramid pierce the 
given plane. 2. Connect the points thus 
obtained in their order, thereby determining 
the required intersection between the plane 
and pyramid. 



84 



DESCRIPTIVE GEOMETRY 



Construction. Fig. 158. Given the pyra- 
mid of which lines J., B^ C\ i>, and E are the 
lateral edges, or elements, intersected by plane 
N. The points of piercing, a, 5, c, c?, and e, 
of the elements with plane N have been de- 
termined by the use of the horizontal project- 
ing planes of the elements (Art. 59, page 42), 
and the lines ah^ hc^ cd^ de^ and ea. joining 
these points of piercing in their order, are the 
lines of intersection of plane and pyramid. 

Since all the auxiliary planes used contain 
point ^, the apex of the pyramid, and are per- 
pendicular to H^ they must contain a line 
through I perpendicular to II\ hence, o, the 
point of piercing of this line and plane iV, is 
a point common to all the lines of intersection 
between plane N and the auxiliary planes. 
By observing this fact the work of construc- 
tion can be slightly shortened. 

The true size of the cut section is obtained 
by revolving each of its points into V about 
F'iVas an axis (Art 48, page 31). 

121. To develop the pyramid. 

Principle. If the pyramid be laid on a 
plane and be made to turn on its edges until 



each of its faces in succession has come into 
contact with the plane, that portion of the 
plane which has been covered by the pyramid 
in its revolution will be the development of 
the pyramid. From the above it is evident 
that every line and surface of the pyramid 
will appear in its true size in development. 

Method. 1. Determine the true length 
of each line of the pyramid. 2. Construct 
each face in its true size and in contact with 
adjacent faces of the pyramid. 

Construction. Figs. 158 and 159. The 
true lengths of the elements and their seg- 
ments have been determined by revolving 
them parallel to F", as at J.^, B^^ etc., a^^ 6j, etc.. 
Fig. 158. The edges F, (7, 7, J, and K are 
already shown in their true lengths in their 
horizontal projections, since the plane of these 
lines is parallel to H. Hence, through any 
point Z, Fig. 159, representing the apex, draw 
a line B^ equal in length to By With Z as a 
center and with a radius equal to C^ draw an 
arc of indefinite length. With the end of B 
as a center and with a radius equal to G^^ 
draw an arc intersecting the first in point n^ 



DEVELOPMENT OF PYRAMID 



85 




thus definitely locating lines C and G. The 
other faces are determined in like manner. 

The development of the cut is obtained by 
laying off from /, on its corresponding ele- 
ment, the true lengths of the elements from 
the apex of the pyramid to the cut section, 
and joining the points thus obtained. 

The base and cut section may be added to 
complete the development of the pyramid. 




86 



DESCRIPTIVE GEOMETRY 



122. To determine the curve of intersection 
between a plane and any cone. 

Principle. The problem is identical with 
that of the intersection of a plane with a pyra- 
mid, for a cone may be considered as being a 
pyramid of an infinite number of faces. 

Method. 1. Pass a series of auxiliary 
planes perpendicular to one of the coordinate 
planes and cutting elements from the cone. 
2. Each auxiliary plane, save the tangent 
planes, will cut two elements from the cone 
and a right line from the given plane. The 
intersections of this line with the elements 
give two points in the required curve. 

Construction. Fig. 160. Let it be re- 
quired to determine the curve of intersection 
between plane N and the oblique cone with 
its circular base parallel to H. Pass a series 
of auxiliary cutting planes through the cone, 
containing its apex, ^, and perpendicular to H. 
Plane Xis one such plane which cuts two ele- 
ments, ah and ac^ from the cone, and the line 
B from plane N. B intersects element ae in 
point e and element ah in point /, and these 
are two points of the required curve of inter- 



section between plane iVand the cone. Simi- 
larly determine a sufficient number of points 
to trace a smooth curve. The planes passing 
through the contour elements of each pro- 
jection should be among those chosen. 

As in the case of the pyramid, the work of 
construction may be slightly shortened by 
observing that since all of the auxiliary planes 
are perpendicular to H^ and pass through the 
apex of the cone, they all contain the line 
passing through the apex and perpendicular 
to H. This line pierces plane iV^in t?, a point 
common to all lines of intersection between N 
and the auxiliary planes. 

The true size of the cut section is deter- 
mined by revolving each of its points into V 
about FiVas an axis. 

123. To determine the development of any 
oblique cone. 

Principle. When a conical surface is 
rolled upon a plane, its apex will remain 
stationary, and the elements will successively 
roll into contact with the plane, on which they 
will be seen in their true lengths and at their 
true distances from each other. 



DEVELOPMENT OF CONE 



87 



Construction. Since in Fig. 160 the base 
of the cone is parallel to H^ the true distances 
between the elements may be measured upon 
the circumference of the base ; therefore, to 
develop the conical surface upon a plane, 
through any point «, Fig. 161, representing 
the apex, draw a line ac^ equal in length to 
a'^c^^ the revolved position of element ac^ Fig. 
160. With a as a center, and with a radius 
equal to tlie true length of the next element 
ah^ draw an arc of indefinite length. With c 
as a center, and with a radius equal to cV^ 
draw an arc intersecting the first in k. This 
process must be repeated until the complete 





88 



DESCRIPTIVE GEOMETRY 



development has been found. The accuracy 
of the development depends upon the number 
of elements used, the greater number giving 
greater accuracy. 

The development of the curve of intersection 
is obtained, as in the pyramid, by laying off 
from the apex, on their corresponding ele- 
ments, the true lengths of the portions of the 
elements from the apex to the cut section, 
and joining the points thus found. 

124. To determine the curve of intersection 
between a plane and any cylinder. 

Principle. A series of auxiliary cutting 
planes parallel to the axis of the cylinder and 
perpendicular to one of the coordinate planes 
will cut elements from the cylinder and right 
lines from the given plane. The intersections 
of these elements and lines will determine 
points in the required curve. 

Construction. Fig. 162. Given the ob- 
lique elliptical cylinder cut by the plane iV. 
Plane X is one of a series of auxiliary cutting 
planes parallel to the axis of the cylinder and 
perpendicular to H. Since it is tangent to 



the cylinder, it contains but one element, A^ 
and intersects the given plane in the line Gr. 
Since lines 6r and A lie in plane X, they in- 
tersect in a, one point in the required curve 
of intersection between the cylinder and the 
given plane iV. Likewise plane Z intersects 
the cylinder in elements and i>, and the 
plane iV^in line K, the intersections of which 
with lines O and D are c and t?, two other 
points in the required curve. 

The true size of the cut section has been 
determined by revolving it into V about F7V 
as an axis. 

125. To develop the cylinder. 

Principle. When a cylinder is rolled upon 
a plane to determine its development, all the 
elements will be shown parallel, in their true 
lengths, and at their true distances from each 
other. Since in an oblique cylinder the bases 
will unroll in curved lines, it is necessary to 
determine a right section which will develop 
into a right line, and upon which the true 
distances between the elements may be laid off. 
This line will be equal to the periphery of the 



DEVELOPMENT OF CYLINDER 



89 




right section, and the elements will be per- 
pendicular to it. The ends of these perpen- 
diculars will be at a distance from the line 
equal to the true distances of the ends of the 
elements from the right section. A smooth 
curve may then be drawn through the ends of 
the perpendiculars. 

Method. 1. Draw a right line equal in 
length to the periphery of the right section. 
2. Upon this right line lay off the true dis- 
tances between the elements. 3. Through 
the points thus obtained draw perpendiculars 
to the right line. 4. On these perpendiculars 
lay off the true lengths of the corresponding 
elements, both above and below the right line. 
5. Trace a smooth curve through the ends of 
the perpendiculars. 

CoxsTRUCTiON. Figs. 162 and 163. The 
secant plane JSF ol Fig. 162 has been so chosen 
as to cut a right section from the cylinder, 
that is, the traces of the plane are perpen- 
dicular to the projections of the axis of the 
cylinder (Art. 62, page 44). Element 1) has 
been revolved parallel to V to obtain its true 



90 



DESCRIPTIVE GEOMETRY 




1 



Fig. 163. 



DEVELOPMENT OF CYLINDER 



91 



length Dj, and d" is projected to c?^, thus ob- 
taining the true lengths of each portion of D. 
Since all the elements of a cylinder are of 
the same length, D^ represents the true length 
of each element, and their segments are ob- 
tained by projecting the various points of the 
cut section upon it, as at 5^, (?j, etc. Upon the 
right line dd^ Fig. 163, the true distances 
between the elements, de^ ch^ etc., have been 
laid off equal to the rectified distances d'c\ 
c'h\ etc. Through points c?, e, 5, etc., perpen- 
diculars to line dd have been drawn equal to 
the true length of the elements. The portions 
above and below dd are iequal to the lengths 
of the elements above and below the cut 
section. 

126. If the axis of the cylinder be parallel 
to a coordinate plane, the development may be 
obtained without the use of a right section. 

Fig. 164 represents an oblique cylinder 
with its axis parallel to V and its bases 
parallel to H. The elements are represented 
in vertical projection in their true lengths, 
and in horizontal projection at their true 
distances apart measured on the periphery 



of the base. If the cylinder be rolled upon 
a plane, the ends of the elements will move in 
planes perpendicular to the elements ; there- 
fore, h' will lie on h^'V perpendicular to 5% and 
at a distance from a" equal to a^h^. Likewise 
c^ will lie on c'^c' perpendicular to C"" and at a 
distance from U equal to h^e^^ etc. The figure 
shows but one half the development. 




92 



DESCRIPTIVE GEOMETRY 



127. To determine the curve of intersection 
between a plane and a prism. 

Principle. A prism may be considered as 
a pyramid with its apex at infinity; hence, this 
problem in no wise differs in principle and 
method from that of the pyramid (Art. 120). 

Construction. Fig. 165. Given the prism 
the elements of wliich are JL, B^ C^ and D 
intersected by plane N, Points J, t?, and d 
are the points in which elements B^ C, and 2>, 
respectively, pierce plane iV, determined by 
the use of the vertical projecting planes of the 
elements (Art. 59, page 42). Element A, if 
extended, will in like manner be found to 
pierce plane iVat point a. By connecting ah^ 
he, cd, and da, the required lines of intersec- 
tion are determined. But, since point a does 
not lie on the given prism, only the portions 
of lines ah and da which lie on the prism, i.e, 
eh and df, are required, and plane iV intersects 
the top base of the prism in line ef. 

Points e and / may be obtained by the use 
of the vertical projecting planes of lines B 
and F, in which case point a is not needed. 




Fig. 165. 



THE HELICAL CONVOLUTE 



93 



The construction may be shortened by 
observing that since the vertical projecting 
planes of the elements are parallel, their lines 
of intersection with the given plane are 
parallel. 

The true size of the cut section is deter- 
mined by revolving each of its points into H 
about ITiV as an axis. 

128. To develop the prism. 

Method. Determine the true size of a 
right section and proceed as in the case of 
cylinder (Art. 125, page 88), or revolve the 
prism parallel to a coordinate plane and pro- 
ceed as for the cylinder (Art. 126, page 91). 

129. The helical convolute. 

Fig. 166 illustrates a plane triangle tangent 
to a right-circular cylinder, the base of the 
triangle being equal to the circumference of 
the base of the cylinder. If the triangular 
surface be wrapped about the cylinder, the 
point c will come in contact with the cylinder 
at <?j, a at a^ and the hypotenuse ac will be- 
come a helix having a^c^ for the pitch, and the 
angle acd will be the pitch angle. If the 



right line ac be revolved about the cylinder 
while remaining in contact with, and tangent 
to, the helix,* and making the constant angle 
6 with the plane of the base, it will generate a 
convolute of two nappes. The lower nappe, 
which is generated by the variable portion of 
the line below the j^oii^t of tangency, will 
alone be considered. 




Fig. 166. 



* For the theory and construction of the helix and 
involute curves see page 104 of "Elements of Mechanical 
Drawinsr " of this series. 



94 



DESCRIPTIVE GEOMETRY 



130. To draw elements of the surface of the 
helical convolute. 

Construction. Fig. 167. Let the di- 
ameter, aH^^ and the pitch, aV, of the re- 




quired helical convolute be given. Then will 
(fi^^v^v^v \^Q ^\^Q vertical projection of the helix, 
and the tangents to the space helix will be the 
elements of the convolute. From any point 




h draw a tangent to the helix. Its horizontal 
projection, h'^k^^ will be tangent to the circle 
of the base at 5^ and the length of this pro- 
jection will equal the arc h'^e^i^ ; therefore, k^ 



DEVELOPMENT OF HELICAL CONVOLUTE 



95 



will be the horizontal projection of the hori- 
zontal trace of the element hk^ the vertical 
projection of which Avill be h'^k'". Similarly 
traces of other elements may be found, and 
their locus will be points in the base of the 
convolute, which curve is an involute of the 
base of the cylinder on w^hich the helix is 
described. Any element of the convolute may 
now be obtained by drawing a tangent to the 
circle of the base of the cylinder, and limited 
by the involute of this same circle. 

131. To develop the helical convolute. 

Since this surface is developable (Art. 98, 
page 67), it may be rolled upon a plane on 
which the elements will appear in their true 
lengths as tangents to the developed helix, 
whicli, being a curve of constant curvature, 
will be a circle. The developed surface will 
be an area bounded by this circle and its 
involute. 

The radius of the circle of the developed 
helix is determined in the following manner: 
In Fig. 168 a, ^, and c are points in the helix, 
and ht is a tangent to the helix at 5, which 
point is equally distant from a and c. The 



projection of these points on to the plane of 
the right section of the cylinder passed through 
h will be at e and d. ac is the chord of a cir- 
cular arc drawn through a, ^, and (?, approxi- 
mating the curvature of the helix, hi is the 
diameter of this circular arc, and Ich a triangle 
inscribed in its semicircumference, and, there- 
fore, a right triangle. Similarly the triangle 
hdk^ in the plane of the right section, is a right 
triangle. In the triangles hcl and hd.k, he will 
be a mean proportional between fh and hi, and 
hd will be a mean proportional between /5 and 
hk. Substituting R for the radius of curva- 
ture of the helix, — , and r for the radius of 
2 

the circle which is the projection of the helix, 

ok — 2 ■ — ^ 

-— , we may obtain he = fh x 2 R and hd7 —fh 

X 2r. Dividing the second by the first, 

=— = — , but — - = cos /3, wdien ^ is the angle 
he R io 

which the chord he makes with the horizontal 



plane; hence, R = 



cos2/5 



As points a and c 



96 



DESCRIPTIVE GEOMETRY 



approacli each other, the chord he will ap- 
proach ht^ the tangent to the helix at 5, and 
at the limit the angle 13 will equal 6^ the angle 
which the tangent makes with the horizontal 

plane, so that R = 



cos2 6> 

This value may be graphically obtained in 
the following simple manner: Fig. 169. Draw 
a tangent to the helix at h. From its inter- 
section with the contour element of the cyl- 
inder, at ^, draw a horizontal line terminated 
by the perpendicular to the tangent through 
h. Then do will be the required radius for 



ac 



the developed helix, since he = -, and he = ed 



cos 



d' 



COS d\ hence, C(i = 



ac 



cos^ 6 cos^ 6 



^ d}a 



Fig. 169. 



Having determined the radius of curvature 
of the helix, draw the circle and its involute 
to obtain the development of the convolute. 
Fig. 170 is the development of the helical con- 
volute shown by its projections in Fig. 167. 




132. To determine the curve of intersection 
between a plane and a surface of revolution. 

In the following cases the axes of revolution 
are considered to be perpendicular to one of 
the coordinate planes. 



INTERSECTION OF PLANE WITH SURFACE OF REVOLUTION 



97 




Method. 1. Pass a series of auxiliary cut- 
ting planes perpendicular to the axis of revo- 
lution. These planes will cut the surface in 
circles, and the given plane in right lines. 2. 
The intersections of the circles and right lines 
are points in the required curve of intersection. 

Construction 1. Fig. 171. Given the 
ellipsoid with its axis perpendicular to V 
intersected by the plane aS'. Plane F is one 
of a series of auxiliary planes perpendicular to 
the axis of the ellipsoid. Then RY is parallel 
to GL; it cuts from the ellipsoid a circle which, 
in horizontal projection, coincides with ITY, 
and in vertical projection is a circle, and it cuts 
from the given plane aS' the line B, intersect- 
ing the circle at i and ^, two points in the 
required curve of intersection between the 
ellipsoid and the given plane. Repeat this 
process until a suf^cient number of points is 
determined to trace a smooth curve. 

It is convenient to know at the outset be- 
tween what limits the auxiliary planes parallel 
to R should be passed, thus definitely locating 
the extremities of the curve. This may be 
accomplished by passing the meridian plane 



98 



DESCRIPTIVE GEOMETRY 



Z7, perpendicular to jS^ intersecting S in line 
D upon which the required points a and b are 
found. It will be noticed that D'' is an axis 
of symmetry of the vertical projection of the 
curve. 

It is also important that auxiliary planes be 
so chosen as to determine points of tangency 
with contour lines. Two such planes, which 
should be used in the case of the ellipsoid, are 
the plane X, which cuts the surface in its 
greatest parallel and contour line in vertical 
projection, and plane TFJ which cuts the sur- 
face in a principal meridian. The former of 
these determines the points of tangency, e" and 
f\ in vertical projection, and the latter defines 
similar points c^ and d^ in horizontal projection. 
Such points as a, ^, c, c?, e, and / are designated 
as critical points of the curve. 

Construction 2. Fig. 172. Given the 
torus with its axis perpendicular to IT inter- 
sected by plane JSf. The auxiliary planes are 
parallel to IT^ and the critical points are a, 5, 
on plane U; e^f on plane JT; and c, d on plane 
W. Also the highest points, ^, z, and the low- 
est points, Z, k, should be determined. 




Fig. 172. 



CHAPTER VI 



INTERSECTION OF SURFACES 



133. Whenever the surfaces of two bodies 
intersect, it becomes necessary to determine 
the line of their intersection in order to illus- 
trate and develop the surfaces. The charac- 
ter of these lines, which are common to both 
surfaces, is determined by the nature of the 
surfaces, and by their relative size and posi- 
tion. The principles involved in the determi- 
nation of these lines, their projections, and 
the development of the intersecting surfaces, 
are fully treated in the chapter on the inter- 
section of planes and surfaces; but it is neces- 
sary to consider the character of the auxiliary 
cutting planes and the methods of using them 



in order to cut elements from two surfaces 
instead of one. 

1 34. Character of Auxiliary Cutting Surfaces. 
The auxiliary cutting surfaces have been re- 
ferred to as planes, but cylinders and spheres 
are also used whenever they will serve to cut 
the intersecting surfaces in right lines or 
circular arcs. 

The character of the auxiliary plane, or sur- 
face, and tlie method of using it, is dependent 
upon the nature of the intersecting surfaces, 
since it is most desirable, and generally possi- 
ble, to cut lines from the surfaces, which shall 
be either rig-ht lines or circular arcs. 



99 



100 



DESCRIPTIVE GEOMETRY 



The following cases illustrate the influence 
of the type of the intersecting surfaces on the 
character and position of the auxiliary cutting 
planes or surfaces : 

Case 1. Cylinder and cone with axes 
oblique to the coordinate plane : Use auxiliary 
planes containing the apex of the cone and 
parallel to the axis of the cylinder. 

Case 2. Two cylinders with axes oblique 
to the coordinate plane : Use auxiliary planes 
parallel to the axes of the cylinders. 

Case 3. Cone and cone with axes oblique 
to the coordinate planes: Use auxiliary planes 
containing the apices of the cones. 

Case 4. A single and a double curved sur- 
face of revolution with parallel axes Avliich are 
perpendicular to a coordinate plane : Use 
auxiliary planes perpendicular to the axes. 

Case 5. A single and a double curved sur- 
face of revolution, two single-curved surfaces 
of revolution, or two double-curved surfaces 
of revolution, with axes oblique to each other 
but intersecting, and parallel to one of the 
coordinate planes : Use auxiliary cutting 



spheres with centers at the intersection of the 
axes. 

Case 6. A double-curved surface of revo- 
lution with axis perpendicular to a coordinate 
plane, and any single-curved surface : If the 
single-curved surface be a cylinder, use auxil- 
iary cutting cylinders with axes parallel to 
that of the cylinder, intersecting the axis of 
the double-curved surface, and cutting circles 
therefrom. If the single-curved surface be a 
cone, use auxiliary cones with apices common 
with that of the given cone and cutting the 
double-curved surface in circles. 

Case 7. A prism may be substituted for 
the cylinder, or a pyramid for the cone, in 
each of the above cases, save Case 6. 

Case 8. Prisms and pyramids, two of the 
same or opposite kind : Auxiliary planes not 
required. Determine the intersection of the 
edges of each with the faces of the other. 

135. To determine the curve of intersection 
between a cone and cylinder with axes oblique 
to the coordinate plane.* 

* It is customary to represent these surfaces with one or 
both of the bases resting on a coordinate plane. 



INTERSECTION OF CONE AND CYLINDER 



101 



Principle. Since the curve of intersection 
must be a line common to both surfaces, it 
will be drawn through the points common to 
intersecting elements. A series of cutting 
planes passed through the apex of the cone 
and parallel to the axis of the cylinder will 
cut elements from both the cone and the cyl- 
inder, and since these elements lie in the same 
plane, they will intersect, their point of inter- 
section being common to both surfaces, and 
therefore a point in the required curve. 

Method. 1. Draw a line through the 
apex of the cone and parallel to the axis of 
the cylinder. 2. Determine its trace in the 
plane of the bases of the cylinder and cone. 
3. Through this trace draw lines in the 
plane of the bases of cylinder and cone cutting 
these bases. These lines will be the traces of 
the auxiliary cutting planes. 4. From the 
points of intersection of these traces with the 
bases of cone and cylinder draw the elements 
of these surfaces. 5. Draw the required curve 
of intersection through the points of intersec- 
tion of the elements of cylinder and cone. 




102 



DESCRIPTIVE GEOMETRY 



Construction. Fig. 1T4. Through the 
apex of the cone draw line A parallel to the 
axis, or elements, of the cylinder. This line 
will be common to all auxiliary cutting planes, 
and its horizontal trace, b'\ will be a point 
common to all their horizontal traces. JTNis 
one such trace which cuts, or is tangent to, 
the base of the cylinder at t?^ and cuts the 
base of the cone in d^ and e'\ Since these are 
points in elements cut from the cylinder and 
cone by the auxiliary plane iV, the horizontal 
and vertical projections of the elements may 
be drawn. Line U will be the element cut 
from the cylinder, while da and ea are tlie 
elements cut from the cone. The intersection 
of these elements at/ and ^ will be points com- 
mon to the cylinder and cone ; hence, points 
in the required curve of intersection. 

The planes iV and jS are tangent planes, the 
former to the cylinder and the latter to the 
cone ; therefore, they will determine but two 
points each. Intermediate planes, such as Jf, 
will cut four elements each, two from each 
surface, and will determine four points of in- 
tersection. Draw as many such planes as may 




Fig. 174. 



INTERSECTION OF CONE AND CYLINDER 



103 



be necessary to determine the required curve. 
Since it is desirable to determine the points of 
tangency between the curve and contour 
elements, strive to locate the intermediate 
planes so that all such elements may be cut by 
the auxiliary planes. 

136. Order and Choice of Cutting Planes. The 
tangent planes should be drawn first, thus 
determining limiting points in the curve, such 
as/, g, k^ Z, and, as will be shown in Art. 137, 
determining whether there be one or two 
curves of intersection. Next pass planes cut- 
ting contour elements in both views, iff is 
one such plane cutting a contour element in 
the horizontal projection of the cylinder and 
determining points i and ?, limiting points in 
the horizontal projection of the curve. 

137. To determine if there be one or two 
curves of intersection. It is possible to deter- 
mine the number of curves of intersection 
between the cylinder and cone previous to 
finding the required points in the line, or lines, 
of intersection. In Fig. 174 it will be observed 
that there is but one continuous curve of 



intersection, and this is due to the fact that 
only one of the two tangent auxiliary planes 
that might be drawn to the cylinder will cut 
the cone. If the surfaces had been of such 
size, or so situated, that the two planes tan- 
gent to the cylinder had cut the cone, as in- 
dicated by Fig. 175, in which only the traces 
of the cylinder, cone, and planes are drawn, 




there would have been two curves, the cylinder 
passing directly through the cone. Again if 
two planes tangent to the cone had cut the 
cylinder, as in Fig. 176, the cone would have 
pierced the cylinder, making two independent 
curves of intersection. The condition shown 
by Fig. 177 is that of the problem solved, 
and indicates but one curve of intersection. 



104 



DESCRIPTIVE GEOMETRY 



138. To determine the visible portions of the 
curve. 1. A point in a curve of intersection 
is visible only when it lies at the intersection 
of two visible elements. 2. The curve of 
intersection being common to both surfaces 
will be visible in vertical projection only as 
far as it lies on the front portion of both sur- 
faces. 3. The horizontal projection of the 
curve is visible only as far as it lies on the 
upper portion of both surfaces. 4. The point 
of passing from visibility to invisibility is al- 
ways on a contour element of one of the sur- 
faces. Fig. 173 represents the cone and 
cylinder with only the visible portions shown. 

139. To determine the curve of intersection 
between two cylinders, the axes of which are 
oblique to the coordinate planes. 

Principle. Auxiliary planes parallel to 
the axes of both cylinders will cut elements 
from each, and their intersections will deter- 
mine points in the required curve. Since the 
auxiliary planes are parallel to the axes, they 
will be parallel to each other, and their traces 
will be parallel. 



Method. 1. Through one of the axes pass 
a plane parallel to the other axis (Art. 74, 
page 52), and this will be one of the auxiliary 
planes to which the other cutting planes will 
be parallel. 2. Determine if there be one or 
two curves of intersection. 3. Beginning 
with one of the tangent planes pass auxiliary 
planes through the contour elements of each 
view, using such additional planes as may be 
necessary to determine the requisite number 
of points in the curve. 4. Draw the curve, 
.determining the visible portions by Art. 138, 
page 105. 

140. To determine the curve of intersection 
between two cones, the axes of which are 
oblique to the coordinate planes. 

Principle. Auxiliary planes which contain 
the apices of the cones will cut elements from 
each of the surfaces. Hence, all the cutting 
planes will contain the line joining the apices 
of the cones, and all the traces of the auxiliary 
planes will intersect in the traces of this line. 

The solution of this problem is similar to 
that of Art. 135, page 100. 



INTERSECTION OF ELLIPSOID AND CYLINDER 



105 



141. To determine the curve of intersection 
between an ellipsoid and an oblique cylinder. 

Principle. Auxiliary cylinders with axes 
parallel to that of the oblique cylinder and 
intersecting the axis of the ellipsoid may be 
chosen of such section as to cut circles from 
the ellipsoid and elements from the cylinder. 

Construction. Fig. 178. Draw any 
parallel Jfc"" in vertical projection and con- 
ceive it to be the horizontal section of an 
auxiliary cylinder, the axis of which is de. 
This auxiliary cylinder will intersect the 
horizontal coordinate plane in a circle having 
e^ for its center and a diameter equal to 5V, 
since all horizontal sections will be equal. 
Points / and k lie at the intersection of the 
bases of the given and auxiliary cylinder, and, 
therefore, points in elements common to both 
cylinders. The intersections of these elements 
with the parallel cut from the ellipsoid, I and 
0, will be points common to the two surfaces 
and, therefore, points in the required curve. 
Similarly determine the necessary number of 
points for the drawing of a smooth curve. 




Fig. 178 



106 



DESCRIPTIVE GEOMETRY 



142. To determine the curve of intersection 
between a torus and a cylinder, the axes of 
which are perpendicular to the horizontal coor- 
dinate plane. 

Principle. Auxiliary planes perpendicular 
to the axes will cut each of the surfaces in 
circles the intersections of which will deter- 
mine points common to both surfaces ; or, 
meridian planes of the torus will cut elements 
from the cylinder and meridians from the 
torus, the intersections of which will be points 
in the required curve. 

Method. 1. Determine the lowest points 
in the curve on the inner and outer surfaces 
of the torus by a meridian cutting plane con- 
taining the axis of the cylinder. 2. Deter- 
mine the points of tangency with the contour 
elements of the cylinder by meridian cutting 
planes containing said elements. 3. Deter- 
mine the highest points in the curve by an 
auxiliary plane cutting the highest parallel. 
4. Determine intermediate points by auxiliary 
planes cutting parallels from the torus and 
circles from the cylinder. 

Construction. Fig. 179. Through the 



axis of the cylinder pass the meridian plane of 
the torus, cutting elements from the cylinder 
and a meridian from the torus. Revolve this 
plane about the axis of the torus until it is 
parallel to V. In this position the vertical 
projections of the elements will be Al and B^ 
and their intersections with the circle cut from 
the torus will be at cl and dl. In counter- 
revolution these points will fall at <?" and c?", 
thus determining the lowest points of the 
curve on the inner and outer surface of the 
torus. The meridian planes iV and M will cut 
contour elements from the cylinder, and the 
vertical projections of their points of intersec- 
tion with the torus at e" and/" will be deter- 
mined as in the previous case. In this manner 
all the points of intersection may be found ; or, 
planes cutting parallels from the torus may be 
used, as H and S^ the former of which is tan- 
gent to the upper surface of the torus cutting 
it and the cylinder in circles which intersect 
at points k and I. The remaining points 
necessary to the determination of the curve 
may be similarly found as shown by the auxil- 
iary plane S. 



INTERSECTION OF TORUS AND CYLINDER 



107 




Fig'- 179. 



108 



DESCRIPTIVE GEOMETRY 



143. To determine the curve of intersection 
between an ellipsoid and a paraboloid, the axes 
of which intersect and are parallel to the ver- 
tical coordinate planes. 

Pkinciple. Auxiliary spheres having their 
centers at the intersection of the axes of the 
surfaces of revolution will cut the intersecting 
surfaces in circles, one projection of which will 
be right lines. 

Construction. Fig. 180 illustrates this 
case. It will be observed that the horizontal 
projection of the parabola is omitted since the 
curve of intersection is completely determined 
by the vertical projections of the two surfaces 
and the horizontal projections of the parallels 
of the ellipsoid. 




Fig. 180. 



CHAPTER VII 



WARPED SURFACES 



144. Warped Surfaces* are ruled surfaces, 
being generated by the motion of a right line, 
the consecutive positions of which do not lie 
in the same plane. The right-line generatrix 
is governed in two distinct ways : 

1. By contact with three linear directrices. 

2. By contact with two linear directrices 
while maintaining parallelism with a plane 
or other type of surface. If the governing 
surface is a plane, it is called a plane director; 
if a cone, it is called a cone director, and the 
generatrix must always be parallel to one of 
its elements. 

* The classification of surfaces considered in Chapter 
III, and especially that portion of the subject relating to 
warped surfaces, Arts. 100 and 101, page 68, should be 
reviewed previous to studying this chapter. 



The following types will be considered, the 
first two being employed to set forth the char- 
acteristic features of this class of surfaces: 

A surface having its generatrix governed 
by three curvilinear directrices. 

A surface having its generatrix governed 
by two curvilinear directrices and a plane 
director. 

An hyperbolic paraboloid, illustrating a sur- 
face the generatrix of which is governed by 
two rectilinear directrices and a plane director. 

The oblique helicoid, illustrating a surface 
the generatrix of which is governed by two 
curvilinear directrices and a cone director. 

The hyperboloid of revolution of one nappe, 
illustrating a surface which may be generated 
by several methods, and the only warped sur- 
face which is a surface of revolution. 



109 



no 



DESCRIPTIVE GEOMETRY 



145. Having given three curvilinear direc- 
trices and a point on one of them, it is required 
to determine the two projections of the element 
of the warped surface passing through the given 
point. 

Principle. Right lines drawn from a given 
point on one directrix to assumed points on 
a second directrix will be elements of a coni- 
cal surface. If the intersection between this 
conical surface and a third directrix be ob- 
tained, it will be a point of an element of the 
auxiliary cone, which is also an element of the 
warped surface, since it will be in contact with 
the three directrices. 

Method. 1. Assume points on one direc- 
trix and draw elements of an auxiliary cone 
to the given point. 2. Determine the inter- 
section between this auxiliary cone and the 
third directrix. 3. Through the given point 
and the point of intersection of the curve and 
directrix draw the required element. 

Construction. Fig. 181. A, B, and (7 are 
three curvilinear directrices of a warped sur- 
face. It is required to draw an element of 



the surface passing through point d on A. 
Assume points on one of the other directrices, 
in this case (7, and through these points, g, /, 
^, and Z, draw lines to d. Since C' is a curved 
line, these lines will be elements of a cone. 
To find the intersection between directrix B 
and the auxiliary cone, use the auxiliary cylin- 
der which horizontally projects B. B^ will be 
its horizontal trace and the horizontal projec- 
tion of the curve of intersection between the 
auxiliary cone and cylinder. Project the 
horizontal projections of the points of in- 
tersection between the elements of the cone 
and cylinder, ?^'*, w^, r^ and s^, to obtain points 
in the vertical projection of the curve of inter- 
section, Ti", m", r", and s", thereby determining 
point 0, the intersection of the directrix B with 
the auxiliary cone. This point must lie on 
the auxiliary cone since it lies on the curve 
of intersection between the cylinder and cone; 
hence, an element drawn through (?,will inter- 
sect (7, the directrix of the cone, and it will be 
an element of the warped surface because it is 
in contact with the three directrices A^ B^ and 



WARPED SURFACE 



111 




Fig, 181. 



C\ doj is, therefore, the required element of 
the warped surface. 

146. Having given two curvilinear directrices 
and a plane director, to draw an element of the 
warped surface. 

Case 1. In which the element is required 
to be drawn through a given point on one of 
the directrices. 

Principle. If a plane be passed through 
the given point parallel to the plane director, 
it will cut the second directrix in a point 
which, if connected with the given point, 
will define an element of the surface, it being 
parallel to the plane director and in contact 
with both directrices. 

Method. 1. Through any point in the 
plane director draw divergent lines of the 
plane. 2. Through the given point of the di- 
rectrix draw parallels to the assumed lines on 
the plane director. 3. Determine the inter- 
section between the plane of these lines and 
the second directrix by use of the projecting 
cylinder of this directrix. 4. Connect this 
point with the given point. 



112 



DESCRIPTIVE GEOMETRY 



CoNSTRUCTiOK. Fig. 182. A and B are 
the directrices, iV the plane director, and d the 
given point. From any point c, in the plane 
director, draw divergent lines E, F, and Gr, 
Through point d draw dt, ds, and dr parallel 
to the lines in the plane director, thus defin- 
ing a plane parallel to iV^. The vertical pro- 
jection of the curve of intersection between 
this plane and the horizontal projecting cylin- 
der of B will be r^'sT. The intersection be- 
tween this curve and B is at m, and dm will 
be the required element of the warped surface. 

147. Case 2. In which an element is re- 
quired to be drawn parallel to a line of the 
plane director. 

Peinciple. If an auxiliary cylinder be 
used which has one of the curved directrices 
for its directrix, and its elements parallel to 
the given line, it will have one element which 
will intersect the second directrix. Such an 
element will be parallel to the given line on 
the plane director, and in contact with both 
directrices ; hence, an element of the warped 
surface. 



Method. 1. Through assumed points on 
one directrix draw lines parallel to the given 
line in the plane director, thus defining an 
auxiliary cylinder. 2. Determine the curve 
of intersection between this cylinder and one 
of the projecting cylinders of the second 
directrix. 8. Through the intersection of 
this curve with the second directrix draw the 
required element parallel to the given line. 

Construction. Fig. 183. A and B are 
the directrices, iV the plane director, and C 
the given line in the plane. e, /, k, and I 
are the assumed points on directrix A through 
which the elements of a cylinder are drawn 
parallel to line C. This auxiliary cylinder 
through A will intersect the horizontal pro- 
jecting cylinder of B in a curve of which aS'" is 
the vertical projection, m is a point common 
to the auxiliary cylinder and the directrix B, 
and dm the required element of the warped 
surface. 

148. Modifications of the two types of sur- 
faces in Arts. 145 and 146 may be made to 
include all cases of warped surfaces. 



WARPED SURFACE 



113 




Fig. 182 



Fig. 183. 



114 



DESCRIPTIVE GEOMETRY 



In the first type the three Imear directrices 
may be curvilinear, rectilinear, or both curvi- 
linear and rectilinear. In the second type, 
with two linear directrices and a plane director, 
the directrices may be curvilinear or rectilinear, 
and a cone may be substituted for the plane. 
All conceivable ruled surfaces may be gener- 
ated under one of these conditions. 

149. The Hyperbolic Paraboloid. If the case 
considered in Art. 146 be changed so that the 
two directrices be rectilinear, while the genera- 
trix continues to be governed by a plane direc- 
tor, the surface will be an hyperbolic paraboloid. 
It is so called because cutting planes will in- 
tersect it in hyperbolas or parabolas. Figs. 
184 and 185 illustrate this surface. In Fig. 
184, A and B are the directrices, and IT the 
plane director of the surface. The positions 
of the generatrix, or elements, are shown by 
the dotted lines. As the elements will divide 
the directrices proportionally, they may be 
drawn by dividing the directrices into an 
equal number of parts, and connecting the 
points in their numerical order. 



This surface is capable of a second genera- 
tion by conceiving the elements D and to be 
directrices, and P to be the plane director. 
In this case the directrices O'and D are divided 
proportionally by the elements which are now 
parallel to P. 

Again, we may consider the lines A^ E, and 
B to be three rectilinear directrices governing 
the motion of the generatrix i>, which is 
fully constrained and will describe the same 
surface as before ; but if the three directrices 
were not parallel to the same plane, the char- 
acter of the surface would be changed, and it 
would become an hyperboloid of one nappe. 

An interesting application of this surface to 
practice is found in the pilot, or " cow catcher," 
of a locomotive, which consists of two hyper- 
bolic paraboloids symmetrically placed with 
respect to a vertical plane through the center 
of the locomotive. Figs. 186 and 187 illus- 
trate the types which are commonly used. In 
the former the plane director is vertical and 
parallel to the rails, and in the latter it is 
horizontal. 



HYPERBOLIC PARABOLOID 



115 





Fig. 186. 




Fig,. 187. 



116 



DESCRIPTIVE GEOMETRY 



150. Through a given point on a directrix, to 
draw an element of the hyperbolic paraboloid. 

Principle. The required element must lie 
in a plane containing the given point and par- 
allel to the plane director. A second point in 
this line will lie at the intersection of the 
second directrix with the auxiliary plane passed 
through the given point. 

Method. 1. Through the given point 
draw two lines, each of which is parallel to a 
trace of the plane director (Art. 71, page 50). 
2. Determine the point of intersection of the 
second directrix with the plane of the auxiliary 
lines (Art. 61, page 44). 3. Connect this 
point of intersection with the given point. 

Construction. Fig. 188. A and B are the 
directrices, and n the given point. Through 
n draw U parallel to the horizontal trace of 
plane iV^, and F parallel to the vertical trace. 
These lines will determine a plane parallel to 
i^. Next obtain the intersection of directrix 
B with the plane of lines B and F. This point 
is m, and mn will be the required element. 



151. Having given one projection of a point 
on an hyperbolic paraboloid, to determine the 
other projection, and to pass an element through 
the point. 

Construction. Fig. 189. m^ is the hori- 
zontal projection of the given point which lies 
on the surface of the hyperbolic paraboloid 
having A and B for its directrices, and JV for 
the plane director. Tlu'ough m draw mg per- 
pendicular to H, and determine its intersection 
with the surface, as follows : 

Determine two elements, cd and ef^ near the 
extremities of the directrices (Art. 150), the 
work not being shown in the figure. Divide 
the jDortion of each directrix limited by the 
elements cd and ef into an equal number of 
parts to obtain elements of the surface (Art. 
149, page 114). Pass an auxiliary plane X 
through the perpendicular mg. This will in- 
tersect the elements at k, Z, and r ; the curve 
/S', connecting these points, will be the line of 
intersection between the auxiliary plane X and 
the warped surface. Since the curve aS' and 



HYPERBOLIC PARABOLOID 



117 



tlie perpendicular throagii m lie in the plane 
X, tlieir intersection will be a point common 
to the perpendicular and the warped surface. 
Therefore m^' and m^ will be projections of the 
required point. 

To obtain the required element, pass a plane 




through this point m, parallel to the plane 
director iV, and determine its intersection 
with one of the directrices (Art. 150, page 
118). Through this, and the point m, draw 
the required element. The last operation is 
not illustrated in the figure. 




118 



DESCRIPTIVE GEOMETRY 



152. Warped Helicoids. Suppose the line 
hc^ of Fig. 190, to be revolved uniformly about 
the line cd as an axis while maintaining the 
angle 6 constant, and at the same time com- 
pelled to move in contact with, and uniformly 
along, the axis. All points in the line, save 
that one in contact with the axis, will gener- 
ate helices of a constant pitch, and the sur- 
face generated will be an oblique helicoid. 
The axis, and the helix described by point 5, 
may be considered as the directrices, and the 
generatrix may be governed by a cone which 
is conaxial with the helicoid. The elements 
of this cone will make the angle 6 with H. 

The generatrix may also be governed by 
two helical directrices and a cone director, as 
in Fig. 191. 

Again, it may be governed by three direc- 
trices, which in this case. Fig. 191, may be 
the two helices and the axis. The V-threaded 
screw is the most familiar application of the 
oblique helicoid (Fig. 147, page 69). 

153. If the generatrix be perpendicular to 
the axis, as in Fig. 192, it may be governed 



by directrices similar to the preceding, but 
the cone director will have become a plane 
director, and the surface generated will be a 
right helicoid. This type is illustrated by the 
square-threaded screw (Fig. 146, page 69). 

154. If the generatrix does not intersect 
the axis, as in the preceding cases, a more 
general type will be generated, as shown in 
Fig. 193. In this case the generatrix is gov- 
erned by two helical directrices and a cone 
director, the generatrix being tangent to the 
cylinder on which the inner helix is described. 

155. Hyperboloid of Revolution of one Nappe. 
This is a surface of revolution which may be 
generated by the revolution of an hyperbola 
about its conjugate axis, as illustrated in Fig. 
194. It is also a warped surface in that it 
may be generated by the revolution of a right 
line about an axis which it does not intersect, 
and to which it is not parallel. Furthermore, 
it will be shown that the rectilinear generatrix 
may be governed by three rectilinear direc- 
trices, by three curvilinear directrices, or by 
two curvilinear directrices and a cone director. 



WARPED HELICOIDS 



119 




Fig. 191 



Fig. 192. 



Fig. 193, 



120 



DESCRIPTIVE GEOMETRY 



In Fig. 194 conceive the generatrix cd as 
making the constant angle c^ct'h'' with a hori- 
zontal plane, and revolving about a vertical 
axis through o. Point e will describe the circle 
of the upper base cgl, point d will describe the 
circle of the lower base dmf^ and the point e, 
the nearest to the axis, will describe the circle 
elan, which is called the circle of the gorge. 
All other points of the generatrix will simi- 
larly describe circles, and by drawing these 
parallels of the surface, the meridian line will 
be determined, and is an hyperbola. Thus 
point s, in the generatrix cd^ will be in the 
position t when it lies in the principal me- 
ridian plane, and f will be a point in the ver- 
tical contour, which is an hyperbola. 

156. Through any point of the surface to 
draw an element. If one projection of the 
point be given, draw a parallel of the surface 
through this point to determine the other pro- 
jection. Through the horizontal projection 
of this point draw a tangent to the circle of 
the gorge, and it will be the horizontal pro- 
jection of the required element, the extremi- 




Fig. 194, 



HYPERBOLOID OF REVOLUTION OF ONE NAPPE 



121 



ties of which lie in the horizontal projections 
of the upper and lower bases of the surface. 
Since either extremity may be regarded as 
lying in the upper base of the surface, there 
are two tangents which may be drawn through 
the given point. They will make equal 
angles with the horizontal coordinate plane, 
and intersect at the circle of the gorge. 

157. The Generatrix may be governed by 
Three Rectilinear Directrices. Fig. 194. If two 
elements, ah and cd^ be drawn through point e 
of the circle of the gorge, either may be taken 
as the generatrix of the surface. One is 
known as an element of the first generation, 
and the other as an element of the second 
generation. 

Conceive ah as fixed and cd as the genera- 
trix. In the revolution about the axis, cd 
will at all times intersect ah^ if these lines be 
extended indefinitely. This may be proved 
as follows : If cd be in the position indicated 
by gf, then the horizontal projections of ah 
and gf will intersect in r^. This point will 
be equally distant from the points of tangency 



e^ and k^\ and since ah and gf make equal 
angles with H^ the distances er and kr must 
be equal, and hence, r must be at the inter- 
section of ah and gf. If, then, we conceive 
three elements of the surface, such as cd^ gf^ 
and Im^ and if we conceive ah as the genera- 
trix, it will intersect each of these elements 
and they may be used as directrices. 

Again, if three parallels be the directrices, 
the generatrix will be fully constrained. 

158. The Generatrix may be governed by 
Two Curvilinear Directrices and a Cone Director. 
Fig. 194. Since the elements of the surface 
are parallel to the elements of a cone, having 
the angle d^'e^'h'" for the apex angle, this may 
be used as a cone director, the generatrix be- 
ing also governed by two parallels of the sur- 
face, such as the bases, or a base and the circle 
of the gorge. 

159. The tangent plane to any point of the 
surface is determined by the elements of the 
two generations drawn through this point. 
The plane determined by lines ah and gf will 
be tangent to the surface at point r, Fig. 194. 



122 



DESCRIPTIVE GEOMETRY 



1 60. Through a right line to pass a plane tan- 
gent to any double-curved surface of revolution. 

By the use of the hyperboloid of revolution 
of one nappe as an auxiliary surface, it is pos- 
sible to make a general solution of problems 
requiring the determination of tangent planes 
to double curved surfaces of revolution, as 
follows : 

Principle. If the given right line be re- 
volved about the axis of the double-curved 
surface of revolution, it will generate an hy- 
perboloid of revolution. A plane tangent to 
both surfaces and containing the given line, 
which is an element of ooe of them, will be 
the required plane. Since one, and only one 
meridian plane at a point of tangency will be 
perpendicular to the tangent plane, and as the 
surfaces of revolution have a common axis, it 
follows that one meridian plane will cut a 
line from the tangent plane which will pass 
through the points of tangency on both sur- 
faces and be tangent to both meridian curves. 
This line and the given line will determine 
the tangent plane. 




HYPERBOLOID OF REVOLUTION OF ONE NAPPE 



123 



Method. 1. Draw the principal meridian 
section of the hyperboloid of revolution 
which has the given line for its generatrix. 
2. Draw a tangent to the principal meridian 
sections of both surfaces. 3. Revolve this 
line about the axis of the surfaces until it 
intersects the given line, observing that its 
point of tangency with the hyperbola is a 
point of the given line. 4. Determine the 
plane of this tangent and the given line. 

Construction. Fig. 195. Having de- 
scribed the hyperboloid of revolution with 
the given line A as its generatrix, draw c'|5j 



tangent to the meridian curves. It will be 
the vertical projection of the revolved position 
of a line tangent to both surfaces. In counter- 
revolution this line will intersect the given 
line A at <?, which is the counter-revolved 
position of the point of tangency Cy This 
must be so, since line A is an element of the 
hyperboloid of revolution and must be in con- 
tact with the parallel through c^ Point h^ 
in counter-revolution is at J, and be will be 
a line of the tangent plane. iV will be the 
plane of be and the given line A^ and, there- 
fore, the required tangent plane. 



CHAPTER VIII 



PROBLEMS 



i6i. Directions for solving the Problems. 

The problems are arranged in pairs, allow- 
ing an instructor to assign them alternately, 
inasmuch as it would not be a wise expendi- 
ture of time for a student to solve all of them. 

The problems are designed to be solved 
within margin lines measuring 7 x 10 inches, 
one such plate constituting an exercise. 

The notation of Art. 6, page 4, is to be 
used. The student should remember that the 
correct lettering of every point and line, and 
the observance of the character of lines, is 
as much a part of the solution of the problem 
as is the correct location of point or line. 

The following abbreviations will be used in 
solving the problems : 

V-pr. signifies Vertical Projection. 

Il-pr. signifies Horizontal Projection. 



P-pr. signifies Profile Projection. 

V-tr. signifies Vertical Trace. 

H-tr. signifies Horizontal Trace. 

P-tr. signifies Profile Trace. 

The coordinates of points will be designated 
as follows : 

1st dimension is the perpendicular distance 
to K 

2nd dimension is the perpendicular distance 
to^. 

3rd dimension is the perpendicular distance 
to P. 

Distances above H are -}- , and below H 
are — . 

Distances before V are -}-, and behind V 
are — . 

Measurements from P are 4- and to the 
left. 



124 



PROBLEMS 



125 



162. Problems. The space required for each 3. Construct the H-jjr. and V-pr. of the 

of the first forty-eight problems is 3^ x 5 following points (Art. 7, page 6). 

inches, and the unit of measure is | inch. <2, 6, 2. 5, 5, — 2. c, — 1, —4. cZ, —4, 3. 

1. Required the distance from Hs and the g, 4, 0. /, — 4, 4. k^ 0, — 2. Z, 4, — 3. m, 0, 0. 

Q, of each of the following points (Art. 7, ??, 4, — 4. 



page 6). 



1«* 



T^' 



T/^7* ]9' 



'k" 



T6' 



ia'' 



^^ 



ie* 



I 

I 
_L_ 



ij"!" 



^r 



2. Required the distance from V^ and the 
§, of each of the folloAving points (Art. 7, 
page 6). 



T«' 






T^' 



I 



ia' 



Tf^ 



\k' 



rr 



i/" 



;(/A 



1 

i 



4. Construct the IT-pr. and F'-pr. of the 
following points (Art. 7, page 6). 

«, - 6, - 4. 5, - 4, 4. (?, 5, 4. d, 9, -4. 
e, -4, 6. /, 0, 10. Jc, -5, 0. Z, 0, 0. 



w, — z, 



n, 



5. Fully describe the following lines 
(Arts. 8, 9, 10, 11, 12, pages 7 and 8). 



v<?-^ 



<?^ 



6. Fully describe the following lines 
(Arts. 8, 9, 10, 11, 12, pages 7 and 8). 
c" 



,Bh 




126 



DESCRIPTIVE GEOMETRY 



7. Required , the H-pr. and V-pr. of the 
following lines (Arts. 8, 9, 10, 11, 12, pages 
7 and 8). 

A^ inclined to F", inclined to H^ in 3 Q. 
B, parallel to H, inclined to V, in 2 Q. 
(7, parallel to P, inclined to ^and F, in 1 Q. 
i), perpendicular to F", in 3 Q. 
E, parallel to H^ inclined to V^ in 4 Q. 
F^ inclined to FJ lying in H^ between 2 Q 
and 3 Q. 

8. Required the H-pr. and V-pr. of the 
following lines (Arts. 8, 9, 10, 11,-12, pages 
7 and 8). 

J., inclined to F, inclined to H., in 1 Q. 
J5, inclined to H^ parallel to V^in 3Q. 
6', perpendicular to F, in 3 Q. 
i>, inclined to F, parallel to H^ in 2 Q. 
U, parallel to CrL, in 4 Q. 
F^ inclined to H^ lying in F", between 3Q 
and 4 Q. 

9. Draw the H-pr. and V-pr. of the follow- 
ing lines (Arts. 8 to 15, pages 7 to 10). 

A and B intersecting in 3Q. A parallel to 



Fand inclined to H\ B inclined to Fand H. 

C and B intersecting in 2Q. C perpendic- 
ular to H; B parallel to GL. 

F and F not intersecting. F perpendicular 
to F; F inclined to Fand H. Both in 4 Q. 

10. Draw the H-pr. and V-pr. of the fol- 
lowing lines (Arts. 8 to 15, pages 7 to 10). 

A and B parallel, and inclined to F and H, 
in 3 Q. 

and B intersecting in 1 Q. O inclined to 
Fand H; B parallel to Fonly. 

F and F intersecting in 3 Q. F parallel to 
^and inclined to F; P parallel to GrL. 

11. Required the H-pr., V-pr., and P-pr. 
of the following lines (Arts. 21-23, pages 
14, 15). State the ^'s in which they appear, 
and the direction of inclination (Art. 17, 
page 10). 

' a, - 6, - 4, 8. \ e, 6, 4, 9. 

5, _ 2, - 4, 0. U, 2, - 4, 0. 



ef 



ah 



e, ~ 2, 4, 10. 
/, - 8, 4, 0. 



PROBLEMS 



127 



ah 



yh, - 5, 1, 0. 



12. Required the H-pr., V-pr.^ and P-pr. 
of the followmg Imes (Arts. 21-23, pages 
14, 15). State the ^'s in which they appear, 
and the direction of inclination (Art. 17, 
page 10). 

a, - 2, 6, 7. ( e, 6, 1, 10. 

U,4, 1, 0. 
e, _ 6, - 4, 9. 
./, -2,4, 0. 

13. Required the H-pr. and V-pr. of the 
following triangles (Art. 21, page 14). 

U, -8,-3, 11. [d, -2, 2, 8. 

ahe\ h, -1, - 10, 7. def e, - 2, 9, 4. 
[^,0,-3,0. |/,-2, 2, 0. 

14. Required the H-pr. and V-pr. of the 
following triangles (Art. 21, page 14). 

U, -Q, -Q, 8. id, -1,-1, 11. 

ahch, -1, -2, 6. c?e/Je, 6,4, 0. 
[^,0, -1,0. [/, 4, -2,6. 

15. Three points, a, h, and 6% lie in P, and 
in 3Q. a and b have their V-prs. in the same 



point, and 5 and e have their H-prs. in the 
same point, a is 4 units from V and ^; h 
is 10 units from F", and c is 8 units from H. 
Determine their H-prs.., V-prs. ^ and P-prs. 
(Arts. 21-23, pages 14 and 15), 

16. Three points, a, 5, and c, lie in P, and 
in 3 Q. a and 6 have their H-prs. in the same 
point, and h and c have their V-prs. in the 
same point. « is 4 units from H^ and 8 units 
from V], e is 10 units from H^ and 6 units 
from V. Determine their H-prs. ^ V-prs. ^ and 
P-prs. (Arts. 21-23, pages 14 and 15). 

17. Draw the H-pr.^ V-pr., and P-pr. of line 
ab. a, - 2, 12, 0. 5, 9, - 2, 22. Determine 
the prs. of the following points in ab. 

c^ equidistant from ^and V. 
c?, the H-tr. of the line. 1 J Art. 16, page 10. 
e, the V-tr. of the line. J [ Art. 24, page 16. 
/, the distance from H twice that from V. 
^, in 4 §, 4 units from H. 
In what Q\ does the line appear? 

18. Draw the H-pr.., V-pr., and P-pr. of line 
ab. a, 6, -6,20. 5,-2,11,0. Solve as for 17. 



128 



DESCRIPTIVE GEOMETRY 



19. Make an oblique projection of 17, repre- 
senting Fi H^ and P in their relative positions, 
and the prs. of the line, and points thereon. 

20. Make an oblique projection of 18, repre- 
senting Fi H^ and P in their relative positions, 
and the prs. of the line, and points thereon. 

21. Draw the prs. of line ah lying in P. 
a, - 2, 12. 5, 12, - 4. Solve as for 17, omit- 
ting c and h (Art. 25, page 18). 

22. Draw the prs. of line ah lying in P. 
a, 12, 8. 5, 2, — 8. Solve as for 17, omitting 
c and h (Art. 25, page 18). 

23. Draw the H-pr. and V-pr. of lines A 
and B, having the following traces. Designate 
the §'s in which they appear, if produced 
(Art. 26, page 18). A, H-tr., 6 units behind 
F, and 10 units to the right of V-tr. V-tr.^ 
8 units below IT. B., H-tr.., 5 units before 
F", and 11 units to the right of V-tr. V-tr.^ 
10 units below H. 

24. Draw the H-prs. and V-prs. of lines C 
and 7>, having the following traces. Designate 
the §'s in which they appear if produced 



(Art. 26, page 18). C, V-tr., 4 units above H, 
and 11 units to the right of H-tr. H-tr., 10 
units behind V. i>, H-tr.., 9 units before F", 
and 12 units to the right of the V-tr. V-tr., 
6 units above H. 

25. Draw the H-pr. and V-pr. of lines ap- 
pearing in the following $'s only (Art. 24, 
page 16). J., 1, 4, 3. B, 1, 2, 3. Note. As- 
sume the traces of the lines and proceed by 
Art. 24, page 18. 

26. Solve as for 25. A, 4, i, 2. B, 2, 5, 4. 

27. Solve as for 25. C, 2, 4. i>, i, 3. 

28. Solve as for 25. C, i, 3. i>, 4. 

29. Solve as for 25. E, 4. F, 2, 4. 

30. Solve as for 25. E, 4, 2. F, 5, 2. 

31. Solve as for 25. iT, 5, 4. Z, 5. 

32. Solve as for 25. K, 2, 1. L, 1, 

33. Draw the H-pr. and V-pr. of line ah. 
a^ _1, _6, 14. h, -8, -1, 0. Through 
the middle point of ah draw line C parallel to 
F, and making an angle of 30° with H. Deter- 
mine the traces of the plane of these lines 
(Arts. 27-30, pages 19, 20). 



PROBLEMS 



129 



34. Draw the H-pr. and V-pr. of line ah. 
^, - 6, - 2, 0. h, - 1, 11, 14. Through the 
middle point of ah draw line C parallel to H^ 
and making an angle of 45° witli V. Deter- 
mine the traces of the plane of these lines 
(Arts. 27-80, pages 19 and 20). 

35. Given line ah. a, 3, 3, 10 ; 5, 7, 9, ; 
and point c, 7, 4, 0. Through point c draw a 
line parallel with ah., and determine the traces 
of the plane of these lines (Arts. 27-30, pages 
19 and 20). 

36. Given line ah. a, — 3, — 8, 9 ; 5, 2, 

— 12, ; and point <?, — 6, — 5, 4. Through 
point e draw a line parallel with ah^ and deter- 
mine the traces of the plane of these lines 
(Arts. 27-30, pages 19 and 20). 

37. Given line ah. «, — 1, — 6, 14 ; ^, — 8, 

— 1, ; and point ^, 6, — 1, 7. Determine the 
traces of the plane of these (Art. 33, page 22). 

38. Given line ah. a, — 6, — 2, ; J, — 1, 
11, 14; and point c, 4, — 4, 8. Determine the 
traces of the plane of these (Art. 33, page 
22). 

39. Determine the traces of the plane in 



which points a, 5, and c lie. a, — 8, — 2, 22. 
h, 8, - 2, 0. c, 6, - 9, 14 (Art. 34, page 22). 

40. Determine the traces of the plane in 
winch points a, 5, and c lie. a, — 8, — 4, 16. 
^, - 3, - 8, 8. (?, - 6, - 12, (Art. 34, page 
22). 

Note. In the following problems the traces 
of the planes are parallel to, or make angles of 
15°, or its midtijDle, with CrL. This angle 
may be determined by inspection. 

41. Draw a triangle on plane M. 
Assume one projection and proceed 
as in Art. 35, page 22. 

42. Draw a triangle on plane S. 
Assume one projection and proceed 
as in Art. 35, page 22. 

43. Determine the prs. of a point 
on plane N which is 6 units from V 
and H (Arts. 36, 37, page 24). 
Through this point draw three lines 
on the plane as follows : J., parallel 
to H; B, parallel to F; C, oblique 
to V and ^ (Arts. 27, 28, page 19). 




vs 



10 




130 



DESCRIPTIVE GEOMETRY 




44. Determine the prs. of a point on plane 
N which is 7 units from V and 4 units from 
^(Arts. 36, 37, page 24). Through this point 
draw three lines on the plane as follows: A^ 

parallel to VN; B, parallel to HN \ 
(7, through a point on VN 12 
units from GL (Arts. 27, 28, 
page 19). 

45. Determine the prs. of a point on plane 
S which is 6 units from V and H (Arts. 36, 

37, page 24). Through this point 

\^ draw three lines on the plane as 

>^ follows: J., passing through 2 Q and 

3 Q\ B, passing through 3 Q and 4 

Q ; (7, passing through 2 Q^ 3 Q^ and 4 Q 

(Arts. 27, 28, page 19). 

46. Determine the prs. of a point on plane 



S which is 7 units from P^and 4 units from H 
(Arts. 36, 37, page 24). Through this point 
draw three lines on the plane as .a^ 

follows: A^ parallel to V\ B^ paral- <^/ 
lei to H \ C, passing through 2 Q^ 3 / 
Q, and 4 Q (Arts. 27, 28, page 19). 

47. Determine the prs. of the following 
points on plane iV, but not lying in 

a profile plane, a, 6 units from I^in '"^ 

3 Q \ 5, 4 units from R in 3 Q ; c, 2 

units from V in 4 Q (Arts. 36, 37 
page 24). 

48. Determine the prs. of the following 
points on plane i^, but not lying in ^fj 

a profile plane, a, 4 units from H hr s 
in 3 Q ; ^,2 units from Vin 1 Q ; <?, 

4 units from R in 2 Q (Arts. 36, 37, page 24), 



VN 



Unit of measure, | inch. Space required for each problem, 21 
and from right-hand division line, in heavy type. 



3 inches. Measurements from GL, in light type, 



PLATE 1 





6"' 


10 






a' 


■' 



12 






6" 


r9 1 


a" 


■6 








12 


n 

a 


-2 


b"^ 


U 



10 




16 







Determine the true length of line A. Case 1. (Art. 39, page 27.) Case 2. (Arts. 40, 41, page 



12 




28.) 



Unit of measure, | inch. Space required for each problem, 21x3 inches. Angles between GL and traces of Dl ATC O 
planes, multiples of 15^. Measurements from GL, in light type, and from right-hand division line, in heavy type. • LM I C ^ 



1 

12 / 

/ 1 


2 


\ ^ 

"^ i 
\ !86 


, 4 

/ 1 

/ 19 


20"^^ 1 






HN 5 


HN 6 


/ 7 


/ ^ 


10 


4^0* 


VN 


16 

yN 


Point c lies in H, -4,-2 


Point c lies in N, -2,-4 






9 

1 / 

1 / 


10 

12 / 


11 

7 

141 / 


12 
12 


'2 »"^ 




/^I2 


2^^" 
"« 



Point c lies in plane N. Determine its distance from VN and HN. (Art. 36, page 24; Art. 42, page SO.) 



Unit of measure, | inch. Space required for each problem, 21x3 inches. Angles between GL and traces of Dl ATf Q 
planes, multiples of 15=. Measurements from GL, in light type, and from right-hand division line, in heavy type. • L.r\ I E. O 







HR 



10 




16 



VR 



HR 






-J 


^>< ,:^" 




20 ^vl, 12 




VR 


^ 



Problems 1-6. Line A lies in plane S. Determine its length by revolving it into V and into H, about VS and HS as axes. (Art. 
43, page 31.) 

Problems 7-12. Line B and point e lie in plane R. Determine the distance between them. (Arts. 42, 43, pages 30, 31.) 



Unit of measure, | inch. Space required for each problem, 21x3 inches. Angles between GL and traces of pi ATC A 
nes, multiples of 15°. Measurements from GL, in liffht type, and frona right-hand division line, in heavv tvne. . 



planes, multipl 



light type, and from right-hand division line, in heavy type. 






10 3 





HN 



17 I2IO 4 





I3ill)i [86 
I / ^3 






10 



20 



II 




12 



^< 


Ci/ 


1 


131 


8! 


16 


„l i L 


^\ 


1 


/* 



Determine the true size of the polygon by revolving it into V or H. (Art. 43, page 31.) 
Note. — In Problems 7-12 the traces of the plane of the polygon should first be determined. 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles bet-ween GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 6 




Dravtr the prs. of an equilat- 
eral triangle on plane N. Its 
center is point a, -6, -4. One 
side of the triangle is 7 units 
long and parallel to V. 




Draw the prs. of a regular 
hexagon on plane R. Point b 
is one extremity of a long 
diameter coinciding with line 
A. Side of hexagon is 5 units 
long. 




Draw the prs. of a square 
on plane S. Its center is point 
c, 8, 5. One side is 8 units 
long and at an angle of 60° 
with HS. 




Draw the prs. of a regular 
octagon on plane T. Its cen- 
ter is point d, -8, -5. Its 
short diameter is 8 units long 
and parallel to H. 





yr 


8< 


^^^1 


22| 


10N2 




1 16 


■ 6^ 


-,^ 1 1 




\^rK'l2 






Draw the prs. of a circle 
tangent to lines A and B. Its 
diameter is lO units. 



Draw the prs. of a circle 
tangent to the traces of plane 
M. Its diameter is 1 2 units. 



Draw the prs. of a circle 
tangent to lines A and B. Its 
diameter is 8 units. 



Dra-wr the prs. of a circle 
tangent to the traces of plane 
W, and lying in 3Q. Its di- 
ameter is 16 units. 



Draw the projections of the figures indicated. (Art. 45, page 32,) 



Unit of measure, ^ inch. Space required for each problem, 21x3 inches. Angles between GL. and traces of ni ATC C. 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. ' '-'^ • t. O 







vs 






Plane S is perpendicular to V 






HN 



II 



VN 



I6y 






VS 



Determme the prs. of the line of intersection between planes N and S. (Arts. 49, 50, page 35, Arc. 52, page 37.) 



Unit of measure, | inch. Space required for each problem, 21x3 Inches. Angles bet-wreen GL and traces of pi ATp "J 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 









HS 


9 


VN 





vs 



VN 



9 8 



vs 



HN 



HS 





10 



S passes through 
2Q and 4Q at SO'^wit 





S passes through 
I Q and 3Q at eO'with V 




S passes through 

2Q and 4Qat3 0' 

with H 



Determine the prs. of the line of intersection between planes N and S. 
Problems 1-4. Solve by Case 2. (Art. 51, page 36.) 
Problems 5-12. Solve by Case 3. (Art. 54, page 38.) 



(Arts. 52-56, pages 37-40.) 



Unit of measure, | inch. Space required for each problem, 21 x 3 inches. Angles bet-ween GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 8 














VN 



HN 



-9 12 

■6 



22 



16 




Determine the prs. of the point in which line A pierces plane N. Indicate the prs. of the point by c'^ and c'^. 
page 42.) 



(Arts. 57-59, 



Unit of measure, l 
planes, multiples of 15 


nch. Space required for 
=. Measurements from GL 


Bach problem, 2 
. ill lig^^t type, 


Lx3 inches. Angles between GL and traces of p. .-pp q 
and from right-hand division line, in heavy type. ' *—f^ • C. ^ 


a\y- 


1 






2 


a" 


-9 3 


6"- 


9 4 




12 


b"- 


.9 


a" 


■4 
HN 


12^^^^^^ 2 


Il2 




12 


^^^^r^- 


T^~- 




6*- 2 


fl^ 


-2 


oT^z 


a^^B 


6'^ 6 


a"- 


1-9 


5 






7 

VN 
12 


a"- 


9 8 

HN ^ 


\ !l2 


a"- 


7.Z 


.^^ 


k. 


6^ 


-9 


^ 


6"- 


•8 


b". 


2 
-4 
•6 

8 


b" 
a"- 


-2 

•8 


!20 16 ^12 




^^ ^9 _ 10 


^20 6| |4 


\7 .^ , '2 




s'' / 1 1 


- !22 ||4 81 lek 


1 
1 

1 2 
201 


18 8| 6 


1 1 '^1 

1 2kr;2 

! v^^%<^ 
1 / 1 / ^-■ 

1 / \ / 




1 

1 


D" i '. 


?L 0" L 1 


4' 1 1 1 


1 


^" ^ I 


1 ^^^^'' 1 1 


10 


Z^ 


K ^ 










^9 




<^''^9 1 



Problems 1-8. Determine 
Problems 9-12. Determine 



the prs. of the point in which line ab pierces plane N. (Art. 60, page 42.) 

the prs. of the point in which line A pierces the plane of line B and D. (Art. 61, page 44.) 



Unit of measure, | inch. Space required for each problem, 21x3 inches. Measurements from GL, in light PLATE 10 
type, and from right-hand division line, in heavy type. 



9 I 





io|\ 


\ 










191 


IS 112 


6 




2 ^J 




.^-^ 












9 II 





Problems 1-4. Determine the prs. of the point in which line A pierces the polygon. (Art. 61, page 44.) 
Problems 5-12. Determine the prs. of the points in \^hich line A pierces the object. (Art. 61, page 44.) 



Unit of measure, | inch. Space required for each problem, 2| x 3 inches. Angles between GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 11 



20^ 




1 


\I|2 


2 


«"r8 

!l4 


^ 


6 


3 


17 ^\lO 


4 


\ 1 
\ 1 


1 ^\ 

1 

1 


^ 


^^^ 


1 \ 


^ 




VN 


-8 5 


HN 


S 


h 

a t8 

.4 / 


V 


/ 7 


9^ 

16/12 


8 


1 HN 


|l2 




! 12 




1 


! ^/^ 


1/12 


^ 


1 

1 

1 


0"^ 


J 




12 ^ 


9 


12/ 


/'° 


HS 




-a" 


HS 


12 


1 




b\2 
1 




yk 


12 




-4 


12 













Problems 1-8. Determine the prs. and true length of the line, ab, measuring the shortest distance from point a to plane N. 
(Arts. 62-65, pages 44, 45.) 

Problems 9-12. Determine the prs. of a perpendicular, ab, to plane S at point b on S. Line ab to be 8 units long. (Art. 36, page 
24:Arts. 62-65, pages 44, 45.) 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Measurements from GL, in light PI ATF ^0 
type, and from right-hand division line, in heavy type. 



-3 2- 




This side 45° 



Dlam, 

Circumscribing 

Circle 14 



Z^ 



P3^_ 



15' 



/^^ 



^^w> 









Problems 1, 2, 5-7. Determine the prs. of the shado-ws of the object on itself and on V and H. 
Problems 3, 4. Determine the prs. of the shadow of the chimney on itself and on the roof. 
Problem 8. Determine the prs. of the shadow of the bracket on itself and on V. 



(Arts. 66-69, pages 46-48.) 



Unit of measure, | inch. Space required for eacla problem, 2i x 3 inches. Angles between GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 



\ 






\^ 


a\7 ' 

112 


2 


HN 

VN 


— 9 3 


HN 


\ Il2 


h 






1 






16 






112 

1 
1 

a is 


il2 
VN 




aVs 

i 1 

a''\2 


8 


5 


16 


/ 8 


6 


8 
1 \^« 

20i 16 


2 


7 


20 




8 16 


14 / 

7 


S 


o'is 


1 


10 

6 


9^ 


^ 




c" 


6"t6 
'1e 


9 


||8 


> ",)' 

". *> 


10 


1 h 

.41 '- 


r9 

-SS 

2 
lO 


II 




10 

■7 




31 

l|6 8 






^'^ 


||2 6 




1 d"- 


O 

-6 


O 

f" 


12 

-3 

■8 



Problems 1-6. Determine the traces of the plane containing point a and parallel to plane 
Problems 7-1 2. Determine the traces of the plane containing point b and perpendicular to 



N. (Art. 71, page 50.) 
line C. (Art. 72, page 50.) 



Unit of measure, ^ inch. Space required for each problem, 21x3 inches. Angles between GL and traces of PIATF 14 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 




9^" 8 












^[\ 


■<^ 






18 


\] 


\8 






iOj 


\ 




2k^ 


\^<' 1 




\ 




Problems 1-4. Through point b pass a plane parallel to lines A and B. (Art. 73, page 51.) 
Problems 5-8. Through line A pass a plane parallel to line B. (Art 74, page 52.) 
Problems 9-12. Through line A pass a plane perpendicular to plane N. (Art. 75, page 52.) 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Measurements from GL, in light 
type, and from right-hand division line, in heavy type. 



PLATE 15 





1 


2 




3 


4 


^,.8 

'2^ 




".^•: ":'• 






,5 ^!" 


30j 28 jiSjie 




281 


5| 1 
;i2 14 


7&X 


'8 


27; I8\|4 is 


1 / 

1 y 




\ 15 


12 1 

19 1 A 


'■■■ X 


1 •» / 

I y 




V 




\J 






17 




|/ 


24 










24 


24 


5 


6 


^, 


7 


8 


ix 






19 


a\ 


r 




2r^^l8J ||5 


4 
5 


4 / 1 
I23II9 12 


1 \^< 

34 2d\ 




if vv.^ 


30| \)2 

14^^ I5L^^ 


19 


,3l/'?^I2 


1 X 


a" 
6" 


-2 
1 1 


21 

b 

a 
QQ 

a ^ 


H6 



Determine the prs. and true length of the line measuring the shortest distance between lines A and B. (Art. 79, page 54. 
Note. — The problem is best solved by passing the auxiliary plane through line A. 



Unit of measure, J inch. Space required for each problem, 2i x 3 inches. Angles between GL and traces of qi at-c ^ A 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type, r LA It. 10 



1 


. 


/ 


2 




3 


9k 






/4 


20/ ^||4 14 


20/ lie 


1 


- 


I6]\ 


J4 6 


|22 


12^1 


/s 






>> 


1 


1 

1 
1 
1 






1 
^^2 / 






8^^ \ 




\. 




8^ 


^8 










5 

8^ Hfi 


HH 


/1^ 


6 


f 




/9 7 




48 


^^ 


8 


1 ^^"--v^ 


J. 


y 


-a 




"20 ^^1 


> 


16 


- 


,0/ 


a 


1-2 




a- 


-2^ 


^ 




L 1* 


1 
1 


^'. 


\ 


^ 


12 

3 


Nlf- 


b'- 


12 






^'"■"^•^i^ 1 


y 








•=»: 


^^^-^^ 




6«ie 






1//V ^"^^'s 










u 
















/l9 9 


7rv^^ 


P^' 


10 


^N 


11 




a" 


-9 




12 


Il6 !6 


.K 


^. 


- 


1 


\ 6 




6* 


■2 
12 






1 


16 


6 


18 


^"\ N2 




a'-2 






^^^"-.^^ 1 










^-4.4 












^"^^6 














*"J 


7 







Problems 1-8. Determine the true size of the angle between line A and plane N. (Art. SO, page 54.) 

Problems 9-12. Determine the true size of the angle between line B and V and H, (Art. 81, page 55.) Note.— Letter the angle 
w^ith V as X, and with H as y. 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles belrween GL and traces of 
mes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 17 











33 28 



Problems 1-4. Determine the true size of the diedral angle between planes N and T. (Arts. 83-85 
Problems 5-8. Determine the true sizes of the diedral angles of the objects. (Arts. 83-85, pages 



pages, 57, 
57, 58.) 



Unit of measure, J inch. Space required for each problem, 2^ x 3 inches. Angles between GL and traces of p| A-fC' 1Q 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. ' »-A It. 1 O 


\l4 


2 


\8 


4 


^-^ 


^^ 


J^^ 


7 


6 

12/ 


6 

12/^ 


^ 


7 
8 


. 8 

/ 
16/ 


/ 


^y^ 


^^ 


S 




9 


vs . 10 


11 


I^S , 12 


HS 








^^ c 




"* 


//S ^ 










1 



Determine the angle between plane S and V and H. (Art. 86, page 58.) Note. —Letter the angle w^ith V as x, and w^ith H as y. 



Unit of measure, | inch. Space required for the problem, 7 X lO inches. 



PLATE 19 




rp- 




HIP RAFTER. 

Determine length — down cut- 
heel cut — side cut — top bevel. 

JACK RAFTER. 

Determine side cut. 

PURLIN. 

Determine down cut — side cut- 
angle between face and end. ' 



Determine the bevels, cuts, and lengths of roof members, as above. (Art. 87, page OO, 



inch. Space required for the problem, 7 x lO inches, 



PLATE 20 




Determine angle of cut on top of purlin (A). Bevel on web of purlin (B). Angle bet-w^een plane of -web of hip rafter and purlin, or 
bend of gusset (C). Angle between top edges of gusset (D). (Art, 87, page 61.) 



Unit of measure, i inch. Space required for each problem, 21 x 3 inches. Angles between GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. 



PLATE 21 



1 


^ 2 


eo'with H. 


75°with H. ^ 


45°with H. 


3 with V. 


HN 


7 




5 


7 


/ 7 

7 


8 

HH 




^^^ 60°with H. 


60'with H. 


60'with V. 


30°with V. 


9 


10 


II 


12 


eo'with V. 

45"with H. 


75°with H. 
45°with V. 


30°with H. 
eo'with V. 


QO^with H. 
45^with V. 



Problems 1-8, 
Problems 9-12. 



Determine one position of the missing trace of plane N. 
Determine the traces of plane S making the given angle 



(Arts. 88, 89, page 62.) 
3 with V and H. (Art. 90, page 62.) 



Unit of measure, J inch. Space required for each problem, 7 x lO inches. 



PLATE 22 





In belt transmission one condition must always be obtained, namely : the point on a 
pulley from which the belt is delivered must lie in the mid-plane of the pulley to which 
it is delivered. If the shafts are at right angles and placed as in the figure, they will run 
in the direction indicated, point b lying in the mid-plane of the large pulley, and point 
c lying in the midplane of the small pulley. But if the direction be reversed a guide 
pulley would be necessary to compel the belt to fulfill the above conditions. The point 
at which the direction of the belt is changed by the guide pulley is governed by con- 
venience. Problem I is that of a steering gear and requires one guide pulley. Prob- 
lem 2 illustrates a condition which necessitates two guide pulleys, although but one is 
to be determined. 




Problem 1. Draw the projections of a guide pulley 8 units diameter, 2 units face, and determine the shaft angle with V and H. 
(Art. 45, page 32.) 

Problem 2. Draw the projections of one guide pulley 8 units diameter, 2 units face, and determine the shaft angle with V and H. 
(Art. 4ft, page 32.) 



Unit of measure, | inch. Space required for each problem, 7 x lO inches. 



PLATE 23 



Locate GL 38 units from lower margin line. Locate point b on GL 
16 units from right-hand margin line. 

Draw the prs. of a regular hexagonal pyramid in IQ, resting on a 
plane which makes an angle of 20^ with H and 75° with V. The pyra- 
mid rests on its apex at a point in the plane, 18, 4. Axis of pyramid is 
perpendicular to the p'ane, and is 12 units long. The short diameter of 
the base is parallel to H; the long diameter is 12 units in length. The 
traces of the plane intersect GL in point b. 



Determine ihe shadow of the pyramid on the plane 



Locate GL 38 units from lower margin line. Locate point b on GL 
16 units from right-hand margin line. 

Draw the prs. of a regular hexagonal prism in IQ, resting on a plane, 
the vertical trace of which makes an angle of 15^ with GL, and the 
horizontal trace, an angle of 45^^ with GL. The center of the base of 
the prism is a point of the plane, 17, 4. The sides of the hexagon are 6 
units and two of the sides are perpendicular to the horizontal trace of 
the plane. Altitude of prism is 12 units. The traces of the plane 
intersect GL in point b. 

Determine the shadow of the prism on the plane. 



Locate GL 38 units above lower margin line. Locate point b on GL 
20 units from righi-hand margin line. 

Draw the prs. of a regular pentagonal pyramid in IQ, resting on a 
plane, the vertical trace of which makes an angle of 15° with GL, and 
the horizontal trace, an angle of 45° with GL. The pyramid rests on 
its apex at a point 17, 4. Axis of pyramid Is perpendicular to the plane 
and is 12 units long. Circumscribing circle of base of pyramid is 12 
units in diameter. One side of pentagon to be parallel to H. The 
traces of the plane intersect GL in point b. 

Determine the shadow of the pyramid on the plane. 



Locate GL 38 units from lower margin line. Locate point b on GL 
60 units from right-hand margin line. 

Draw the prs. of a cube in IQ, resting on a plane which makes an 
angle of 30° with H and 65° with V. The center of the base of the 
cube is 10 units from the horizontal trace of the plane and 8 units from 
V. One diagonal of the base is parallel to H. Edge of cube is 8 units. 
The traces of the plane intersect GL in point b. 

Determine the shadow of the cube on the plane. 



Dra-w the prs. of a solid resting on an oblique plane, and determine the prs. of its shadow^ on the plane. 



of measure, J inch. Space required for each problem, 5x7 inches. Measurements from GL, in light type, pt > ^^ OA. 
»m right-hand division line, in heavy type. r l-r^ IC ^■f 



Unit 
and from right 




30 21 14 


















8 


30 


I2T 


10 




1 


5 
u 




9ia 






19 




Problems 1, 2, 8. Draw the traces of a plane which is tangent at point a of the surface. (Arts. 104, 105, pages 72-74.) 

Problems 3, 4, 7. Draw traces of planes -which are tangent to the surface and contain point a. (Art. 106, page 74 ; Art. 108, page 75.) 

Problems 5, 6. Draw^ traces of planes w^hich are tangent to the surface, and parallel to line B. {Art. 107, page 75 ; Art. 109, page 76. ) 



Unit of measure, J incla. Space required for each problem, 5x7 inches. Measurements from GL, in light type, qi aT"C OR 
and from riarht-hand division line, in heavy type. r U.r\ I C ^\J 







23 19 




Vertex and focus of 
parabola are located 




!28 



I 

a '6 







Problems 1-4. Draw the traces of a plane which is tangent at point a of the surface. (Art. Ill, page 78.) 

Problems 5, 6. Draw the traces of a plane w^hich contains point a and is tangent to the surface at the given parallel. (Art. 112, 
page 79.) 

Problems 7, 8. Draw the traces of planes tangent to the sphere and containing line A. (Art. 114, page 80.) 



Unit of measure, | inch. Space required for each problem, 7 x lO inches. Angles bet^veen GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type 



PLATE 26 




Determine the intersection of the plane -with the solid, and develop the surface. 

Problem 1. (Arts. 120, 121, pages 83, 84.) Problem 2. (Art. 127, page 92.) Problem 3. (Arts. 124, 125, page 88.) Problem 4. 
(Arts. 122, 123, page 86.) 



Unit of measure, | inch. Space required for each problem, 7 x lO inches. 



PLATE 27 



The generatrix makes an angle 
of 15=' with H, and one revolution. 




Pitch 12 units: length of longest 
element, !9 units. 



-26- 





Prohlems 1, 2. Draw and develop an helical convolute. (Arts. 129-131, pages 93-95.) 

Problems 3, 4. Delvelop one quarter of the lamp shade. Problem 3. (Art. 126, page 91.) Problem 4. (Art. 123, page 86.) 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles bet-wreen GL and traces of qi K't'^ f\f% 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. ■ L-M I C ^O 







I20 





A is the generatrix of I 
an hyperboloid of revolu- 24 





Determine the intersection of the plane N with the double-curved surface of revolution. (Art. 132, page 96.) 



Unit of measure, l inch. Space required for each problem, 7 x lO inches. Measurements from GL, in light type, pi A-fc OQ 
and from right-hand division line, in heavy type. r l-M I C ^^ 




Determine the intersection between the solids. 



Unit of measure, J inch. Space required for each problem, 7 x lO inches. Measurements from GL, in light type, qi a TC QO 
and from right-hand division line, in heavy type. r L_M I C OU 






37 28 16 








Determine the intersection bet-ween the solids. 



Unit of measure, A inch. Space required for each problem, 5x7 inches. Measurements from GL, in light type, pi A-r-p Q 1 
and from right-hand division line, in heavy type. ' I— M I C O I 





i_ 



•t-^ 








t 


/ t 

^ 




— 




1 2 


(] "1 




CO 


(J 1 


\ 


. T 




B2 1 




7^ 


1 ' 



14 







Determine the intersection between the solids. 



The unit of measurement is 1/1 6 inch. Space required 7 x lO inches. 



PLATE 32 




A sectional and end view of a portion of a boiler is shown. It is required to develop one-half of ihe 
dome, and that part of the slope sheet lying between the elements ef and cd, adding the amount necessary 
for flanging or lap 

Make two full views of the boiler, indicating the thickness of plate by a single line. Determine the 
projection of the two curves of the flange ab, observing that the radial elements of the flange make angles 
varying from 0° to y°. The elements of the slope sheet will be parallel to each other and to the vertical 
plane, thus appearing in their true lengths in vertical projection. The development may be obtained by 
either of the following methods, the first being that used in practice. First method: the elements being 
parallel to V, the development may be obtained as in Art. 126, page 91, the amount indicated for the lap 
being added to the sheet. Second method : obtain a right section VX by Art. 118, page 82, the development of which 
will be a right line, mn. To obtain the distance between the elements revolve the plane of the right section parallel 
to P, and measure the required distances in the profile projection, or end view. Develop as in Art. 125, page '' 



Draw and develop the dome and connection sheet, or slope sheet, of a locomotive boiler. 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles between GL and traces of pi A-yc QQ 
anes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. > *—f^ I *— OO 




Problems 1, 2. Draw an element of the -w^arped surface through point a. (Art. 145, page llO.) 

Problems 3, 4. Draw an element of the warped surface through point a. (Art. 146, page 111.) 

Problems 5, 6. Draw an element of the warped surface parallel to line C of the plane director, N. (Art. 147, page 112.) 

Problems 7, 8. Dravsr an element of the hyperbolic paraboloid through point a, of a directrix. (Art. 150, page 116.) 



Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles bet^veen GL and traces of 
planes, multiples of 15°. Measurements from GL, in light type, and from right-hand divison line, in heavy type. 



PLATE 34 





A is the generatrix of an hyper- 
boloid of revolution. 




A is the generatrix of an hyper- 
boloid of revolution. 




A is the generatrix of an hyper- 
boloid of revolution. 



A is the generatrix of an hyper- 
boloid of revolution. 







Minor Axis 



Problems 1, 2. Dra-w an element of the hyperbolic paraboloid through point a. (Art. 151, page IK 

Problems 3, 4. Dra-w an element through point a. (Art. 156, page 120.) 

Problems 5, 6. Dra-w a plane tangent to the surface at point a. (Art. 159, page 121.) 

Problems 7, 8. Through line A pass a plane tangent to the surface. (Art. 160, page 122.) 



ADVERTISEMENTS 



ELEMENTS OF 

MECHANICAL DRAWING 

Use of Instruments, Geometrical Problems, and Projection 
By GARDNER C. ANTHONY, Sc.D. 

Professor of Drawing and Dean of the Department of Engineering, Tufts College 
Member of the American Society of Mechanical Engineers 

THIS is a text-book rather than a copy-book. It establishes principles and suggests 
methods, but permits freedom in their application. 

The system of projection taught is that which the best practice demands, and exam- 
ples have been selected with a view to establishing its principles with the least expenditure 
of time. The solution of geometric problems is required by practical methods in use by 
draftsmen, as well as by the ordinary geometric construction. 

The methods employed for the representation of objects oblique to the planes of pro- 
jection give a clear and comprehensive understanding of the subject. 

The graphic statement of problem.s, which gives a definite lay-out, is a great labor- 
saving device for instructor and student. 



Revised Edition. Cloth. Illustrated. i6o pages, with ig6 Illustrations and 228 Problems. $1.^0 



D. C. heath & CO., Publishers 
BOSTON NEW YORK CHICAGO 



MACHINE DRAWING 

The Principles of Graphic Expression as illustrated by Machine Drawing and 
Sketching, with Numerous Illustrations of Practical Examples 

By GARDNER C. ANTHONY, Sc.D. 

Professor of Drawing and Dean of the Department of Engineering, Tufts College 
Member of the American Society of Mechanical Engineers 

THIS treatise teaches the appHcation of the principles of projection to the illustration 
of machinery, and the more concise method of graphically expressing mechanical 
ideas. It also informs the student concerning the many exceptions to the laws of projec- 
tion, and furnishes practical examples to serve as problems to students and suggestions to 
draughtsmen. 

The book presupposes a knowledge of the use of instruments and of the theory of 
orthographic projection. It is the advocate of no special system of lining, figuring, or 
lettering, and the plates represent a variety of types in drawing which will be serviceable 
to the draftsman who seeks a terse, accurate, and complete expression of mechanical 
ideas. 

Cloth. 6j pages of Text. i8 Folding Plates with 40 Illustrations. $1.^0 



D. C. HEATH & CO., Publishers 

BOSTON NEW YORK CHICAGO 



ESSENTIALS OF GEARING 

By GARDNER C. ANTHONY, Sc.D. 

Professor of Drawing and Dean of the Department of Engineering, Tufts College 
Member of the American Society of Mechanical Engineers 

THIS treatise comprises the course of instruction and problems given by Professor 
Anthony in college and evening drawing schools for several years past, and is the 
result of his practical experience in connection with designing and constructing gears. 

Besides numerous cuts there are fifteen folding plates illustrating the principles and 
practice of describing gear teeth. A series of progressive problems is given, illustrating 
the principles set forth in the text, and also designed to encourage thorough investigation 
of the subjects by suggesting lines of thought and study beyond the limits of this book. 

A definite lay-out for each problem is given, the necessary instruction for its solution 
is clearly stated, and numerous references to the text require the student to make a care- 
ful study of the subject before performing the problem. This enables a variety of original 
problems to be solved by a class with no additional labor on the part of the instructor. 



Cloth. 84 pages of Text and 75 Folding Plates. $1.^0 



D. C. HEATH & CO., Publishers 

BOSTON NEW YORK CHICAGO 



A TEXT-BOOK OF 

FREEHAND LETTERING 

By FRANK T. DANIELS, A.M.B. 

Formerly Assist a?it Professor of Civil Engineeri?ig i?i Tufts College 

THE plan of this book recognizes the fact that but Httle time can be given in tech- 
nical schools to lettering as a special subject. Exercises are laid out and carefully 
planned, so that work may proceed rapidly. 

Consideration is given to only the plain styles in common use by draftsmen. The 
methods suggested -are wholly freehand, and are believed to be best calculated to lead to 
freedom of execution and excellence of result. 

There is throughout an insistence that good lettering is a matter of design rather than 
the following of arbitrary rules. Copying is discouraged, and in many exercises is made 
impossible. Instead, there is ample exposition of the principles of design necessary to 
guide the student. 

Cloth. y8 pages, with i6 Plates and 12 Figures in the Text. $1.00 



D. C. HEATH & CO., Publishers 

BOSTON NEW YORK CHICAGO 



A TEXT-BOOK OF 

TOPOGRAPHICAL DRAWING 

By FRANK T. DANIELS, A.M.B. 

Formerly Assistant Professor of Civil Engineering in Tufts College 

THE aim of this book is to furnish a concise but thorough course in such drawing 
and calculations, and in the use of such instruments, as are needed in connection 
with topographical engineering, including the design, construction, and examination of 
earthwork. A good course in mechanical drawing is presupposed. 

The book is intended to be especially useful in connection with a course in surveying, 
the work being so arranged and explained that it may precede, accompany, or follow such 
a course. Thus none of a student's time need be lost because of inclement weather or 
other interruption to his field work. 

A notable feature is the introduction of numerous exercises and problems, exemplify- 
ing the text and forming useful supplements to field notes. 



Cloth. 144 pages, with 10 Plates and §6 Figures in the Text. $1.^0 



D. C. HEATH & CO., Publishers 
BOSTON NEW YORK CHICAGO 



APR 3 1909 



